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MC · lesson 9 of 12 · 120 min

Motion and Acceleration

Motion problems reward careful bookkeeping, so this lesson starts with reference points and choosing a positive direction. From there it separates distance from displacement and speed from velocity, works through a return trip and a road-speed conversion, and defines average acceleration with signed examples that show why negative does not always mean braking.

Reading · 120 minVideo · 11 minPractice questions (50)

Video lesson: Motion and Acceleration

Video transcript

0:00 Welcome to this explainer on motion and acceleration. Today, we're going to build your completely calculator free toolkit for the ASVAB exam. Look, if you've ever felt a little shaky with math, or if physics equations just look like a total foreign language to you, don't worry, you are in exactly the right place. We're going to break down these concepts so clearly that you'll literally be solving motion problems entirely in your head. Here's our game plan for today. We're looking at position, distance, and velocity, then introducing and calculating acceleration, constant acceleration formulas, reading motion graphs, special motion and exam traps, and finally we'll finish up with some check yourself questions. Position, distance, and velocity. Pinpointing where you are. Okay, first things first. The absolutely crucial distinction you need to make is between distance and displacement. Distance is your total path. It never goes backward. It just keeps adding up. So it is always, always positive. Displacement, on the other hand, well, that's just the straight line from exactly where you started to where you finished. And importantly, it has a sign, positive or negative, that tells us which direction you went.

1:08 So if you walk out to the mailbox and walk right back to your front door, your total distance might be 100 feet. But your displacement, that's zero, because you ended up exactly where you started. Let's put this into practice using our displacement equation, which is really just final position minus initial position. Let's track a pallet jack. It's pushed 30 meters east, which we're going to call positive 30, then it gets pulled 12 meters back west, which is negative 12. If we want the total distance, we just add those path lengths together. 30 plus 12 is 42 meters. Simple. But for displacement, we have to add the signed moves. Positive 30 plus negative 12 leaves us with positive 18 meters. So it traveled over 42 meters of floor, but it's really only 18 meters away from its starting point. Now, notice how distance and displacement directly affects speed and velocity. Speed uses distance while velocity uses displacement. Let's take this runner. They go 300 meters east in 60 seconds, then turn around and go 100 meters west in 40 seconds. Their total distance is 400 meters, and their total time is 100 seconds.

2:10 400 divided by 100 gives an average speed of 4 meters per second. But what about their velocity? Well, because they double back, their displacement is only positive 200 meters. 200 divided by that same 100 seconds gives an average velocity of only positive 2 meters per second. On the exam, you're definitely going to need to convert units, and this is an awesome chance for some easy mental math. The standard for science is meters per second, and the magic number you've got to remember is 3.6. One meter per second equals exactly 3.6 kilometers per hour. So to convert 72 kilometers per hour into meters per second, just divide by 3.6. 72 divided by 3.6 is a clean 20 meters per second. You can also just rearrange our constant speed equation to easily find travel time. Say a truck travels 150 kilometers at a steady 60 kilometers per hour. Time equals distance divided by velocity. 150 divided by 60, that's 2.5 hours. Introducing and calculating acceleration, changing your velocity. So what exactly is acceleration?

3:14 It's just how fast your velocity is changing. The average acceleration is simply that change in velocity, divided by the time it took. Let's look at this cart. It's rolling west, which we'll call the negative direction, at 6 meters per second. That means its initial velocity is negative 6. Two seconds later, it's rolling west at 2 meters per second, making its final velocity negative 2. We take final minus initial, negative 2 minus negative 6. Remember, subtracting a negative makes a positive, so that gives us positive 4. Divide that by 2 seconds, and we get an acceleration of positive 2 meters per second squared. Now, notice how the acceleration is positive, but the cart is moving west and actually slowing down. We're going to talk a lot more about that exact trap in just a minute. Constant acceleration formulas. Modeling steady changes. Since the ASVAB usually assumes a steady constant change, we have three super reliable formulas we can lean on for these calculator free problems. Our first formula finds velocity after a known amount of time. It just says your final velocity equals your starting velocity plus your acceleration multiplied by time.

4:23 Imagine a forklift starts from rest, meaning starting velocity is zero, and it accelerates evenly at 2.5 meters per second squared for 8 seconds. We plug that in. 0 plus 2.5 times 8. Since 2.5 times 4 is 10, multiplying by 8 gives us an even 20 meters per second. See? Just straightforward mental math. Our second formula is for displacement. I know it looks a little long, but honestly, half of it usually just disappears. It says displacement equals starting velocity times time plus one half acceleration times time squared. Look at this go-kart leaving the starting line from rest. Because that starting velocity is zero, that whole first piece just becomes zero. All we actually care about is the second half. One half times the acceleration of 2 times the time of 5 squared. Well, half of 2 is 1 and 5 squared is 25, so the answer cleanly gives us exactly 25 meters. But what if the test doesn't give you the elapsed time at all? No problem. Use our third formula, which perfectly skips time entirely. For a vehicle stopping, we can rearrange it to effortlessly find braking distance.

5:32 Let's look at a van moving at 25 meters per second that hits the brakes at 5 meters per second squared until it comes to a dead stop. Our formula says braking distance equals the starting velocity squared divided by 2 times the acceleration. 25 squared is 625, 2 times 5 is 10. 625 divided by 10, that's simply 62.5 meters of clear road needed. Reading motion graphs, visualizing the math. Before we even draw a line on a graph, a simple data table can instantly reveal what's going on. Take a look at how velocity rises here. At zero seconds, it's 2. Then it's 5. Then 8. Then 11. It is rising by equal steps, exactly 3 meters per second every single second. This equal gain in equal time is the absolute definitive signal of constant acceleration. Now, when you're looking at a position time graph, the slope of the line is literally your velocity. It's just rise over run. If a line starts at a position of 10 meters at zero seconds and falls straight down to a position of 2 meters at 4 seconds, we can easily find the velocity. The rise is 2 minus 10, which is negative 8.

6:38 The run is 4 minus 0, which is 4. Negative 8 divided by 4 proves a steady velocity of negative 2 meters per second. If you want the displacement, just find the area under the line. A scooter's velocity rises in a straight line from 0 to 8 meters per second over 4 seconds. That shapes a triangle on the graph. Area of a triangle is one half base times height. One half times 4 times 8 is 16 meters. Then, it holds steady at 8 meters per second for 3 seconds. That forms a rectangle. Base times height is 3 times 8, which is 24 meters. Your total displacement is just 16 plus 24, which is 40 meters. Special motion and exam traps. Gravity, curves, and tricks. Alright, let's talk gravity. When an object is in free fall, it accelerates downward at 9.8 meters per second squared. If a bolt falls for 2 seconds, its speed is just gravity times time. 9.8 times 2 is 19.6 meters per second. Its drop distance is one half gravity times time squared. Well, 2 squared is 4, and half of 4 is 2, so we again just multiply 9.8 by 2.

7:47 It hits a speed and a drop distance that are both wonderfully close to 20. Now, what about moving in a circle? Remember this. Because your direction is constantly changing, you are always accelerating. Even if your actual speed is perfectly constant. The formula for this inward centripetal acceleration is velocity squared divided by the radius. A cart taking a 50 meter curve at 10 meters per second has an acceleration of 2. But if we tighten that curve to a 25 meter radius, 10 squared divided by 25 gives us 4. So simply halving your turn radius instantly doubles your acceleration. Okay, listen closely because I'm going to let you in on the common exam traps you absolutely must avoid. First, zero velocity does not mean zero acceleration. Think of a ball thrown straight up in the air. At the very peak, its velocity is zero for a split second. But gravity is still pulling it down at 9.8 meters per second squared. Second, negative acceleration does not automatically mean braking. It just means the acceleration points in the negative direction. The golden rule here is if velocity and acceleration have the same signs you are speeding up.

8:54 If they have opposite signs, you are slowing down. And finally, because speed is squared in our formulas, doubling your speed doesn't just double your braking distance. It literally quadruples it. Check yourself questions, prove your new competence. I'll read the question, give you a brief pause to think, and then we'll break down the answer. Here's question one. A worker walks 25 meters east along a warehouse aisle, then walks 10 meters back west. What total distance does the worker cover? Is it A, 25 meters, B, 35 meters, C, 15 meters, or D, negative 15 meters? Take a second to lock it in. Okay, got it? The answer is B, 35 meters. Remember our rule from the very beginning. Distance simply adds all path lengths together, totally regardless of direction. 25 plus 10 is 35. Let's do one more. Question two. A forklift's top speed is listed as 54 kilometers per hour. What is this speed in meters per second?

9:56 A, 54. B, 15. C, 0.9, or D, 194.4. Think back to our conversion shortcut. All right, the answer is B, 15 meters per second. This is safely found by mentally dividing 54 by our magic science constant of 3.6. Look at that. You are officially equipped with a rock solid calculator free toolkit for any kinematics trap the ASVAB throws your way. You understand the difference between distance and displacement? You know exactly how to read motion graphs? And you can solve constant acceleration formulas using just the power of your own brain. So, review these steps, trust your mental math, and ask yourself, what motion problem will you conquer next?

Contents · 75 sections
  1. Lesson objective and exam scope
  2. Position needs a reference point
  3. Choosing the positive direction
  4. Distance versus displacement
  5. Worked example: a return trip
  6. Speed as a rate of distance
  7. Average velocity uses displacement
  8. Worked example: average speed and velocity
  9. Instantaneous speed and velocity
  10. Units for speed
  11. Worked conversion: road speed
  12. Time and distance at constant speed
  13. Worked example: travel time
  14. Average acceleration defined
  15. Acceleration versus speed
  16. Positive and negative acceleration
  17. When an object slows down
  18. Worked example: signed acceleration
  19. Constant acceleration as a model
  20. Velocity after a known time
  21. Worked example: start from rest
  22. Displacement during constant acceleration
  23. Worked example: acceleration from rest
  24. Average velocity for constant acceleration
  25. Worked example: changing velocity evenly
  26. Distance without elapsed time
  27. Worked example: required stopping distance
  28. Braking distance versus stopping distance
  29. Reaction distance example
  30. Effect of speed on braking distance
  31. Motion graphs: position versus time
  32. Worked example: position graph slope
  33. Motion graphs: velocity versus time
  34. Worked example: area under velocity graph
  35. Signed area and reversing direction
  36. Worked graph: forward then backward
  37. Acceleration-time graph interpretation
  38. Uniform acceleration and graph shape
  39. Free fall as accelerated motion
  40. Worked example: dropped object
  41. Upward throw under gravity
  42. Worked upward throw
  43. Mass independence in ideal free fall
  44. Circular motion and direction change
  45. Worked turn-radius comparison
  46. Relative motion in one dimension
  47. Head-on closing speed
  48. Same-direction closing speed
  49. Two-stage journeys
  50. Worked example: acceleration then cruise
  51. Average speed over unequal periods
  52. When average acceleration is zero
  53. Rounding and significant units
  54. Dimensional analysis of motion formulas
  55. Choosing the right equation
  56. Exam trap: zero speed at one instant
  57. Exam trap: negative means braking
  58. Exam trap: mixing distance and graph area
  59. Exam trap: doubling acceleration or speed
  60. Guided drill: three quick decisions
  61. Worked example: speed is not velocity
  62. Reading a motion table
  63. Instantaneous acceleration from a curve
  64. Reverse direction under constant acceleration
  65. Displacement is not always distance
  66. Deriving the stopping-time relation
  67. Deriving distance using average speed
  68. Why curved-path acceleration matters
  69. Acceleration and mechanical causes
  70. Ideal assumptions and real workshop machines
  71. Graph intersection and pursuit
  72. Acceleration on different coordinate axes
  73. Speed after distance with no stopwatch
  74. Final integrated self-check
  75. Review checklist and next connection
Visual lesson

Lesson slides

15 slides
Motion and Acceleration – slide 1 of 15

Lesson objective and exam scope

After this lesson you should be able to describe where an object is relative to a chosen origin, tell distance from displacement and speed from velocity, and work with constant acceleration using signed values. You will read position, velocity and acceleration graphs, handle free fall and upward throws, and chain two-stage trips. Expect short ASVAB-style calculations where units and signs matter.

Position needs a reference point

A position only means something once you say where it is measured from. That starting mark is the origin. Saying a forklift is "15 m" tells you nothing until you add "15 m east of the loading dock door." Pick the origin once, then measure every later position from that same point so your numbers stay comparable.

Choosing the positive direction

On a straight line you also choose which way counts as positive. If east is positive, a point 8 m west of the origin sits at negative 8 m. The choice is arbitrary, but it must stay fixed for the whole problem. A minus sign on position, velocity or acceleration then simply means "pointing the negative way," not "less than nothing."

Number line: origin 0, east positive, point at x = −8 m (8 m west).

Distance versus displacement

Distance is the total length of path covered, so it is never negative and it keeps growing as long as you move. Displacement is the straight-line change in position, final minus initial, and it carries a sign or direction.

Walk to a mailbox and back and your distance is the whole route, but your displacement is zero.

Worked example: a return trip

A pallet jack is pushed 30 m east along an aisle, then pulled 12 m back west. Take east as positive.

Distance adds the path lengths: 30 m plus 12 m gives 42 m.

Displacement adds signed moves:

m

The jack ends 18 m east of where it started after covering 42 m of floor.

Number line: +30 m east, then −12 m back, ending at +18 m; distance 42 m, displacement +18 m.

Speed as a rate of distance

Average speed tells how fast path length is covered on average. Divide the total distance by the total elapsed time.

A drone that flies a 600 m route in 120 s has an average speed of 5 m/s. Speed has no direction and is never negative.

Average velocity uses displacement

Average velocity divides displacement, not distance, by elapsed time, so it keeps a sign.

Because displacement can be smaller than distance, average velocity can be much smaller than average speed. Any trip that ends where it began has zero average velocity no matter how far or fast the object went.

Worked example: average speed and velocity

A runner goes 300 m east in 60 s, then 100 m west in 40 s. East is positive.

Total distance is 400 m in 100 s, so average speed is 4 m/s.

Displacement is

m

so average velocity is

m/s

which means 2 m/s east.

Instantaneous speed and velocity

Instantaneous values describe one moment rather than a whole trip. A car speedometer shows instantaneous speed: how fast you are going right now. Instantaneous velocity is that same size plus a direction, such as 20 m/s north. Over a very short time interval, the average velocity becomes the instantaneous velocity, which is why the two often agree for steady motion.

Units for speed

Any length unit divided by any time unit is a valid speed unit: m/s, km/h, ft/s and mi/h all work. The science standard is meters per second. Two handy conversions are

and 1 mi/h is about 0.447 m/s. Divide km/h by 3.6 to get m/s; multiply m/s by 3.6 to go back.

Worked conversion: road speed

Convert 72 km/h to meters per second.

Change kilometers to meters and hours to seconds in one line:

That gives

m/s

The shortcut gives the same answer: 72 divided by 3.6 is 20 m/s.

Time and distance at constant speed

When speed does not change, distance grows in direct proportion to time.

Rearrange as needed:

or

Keep units matched. If speed is in m/s, time must be in seconds and distance comes out in meters.

Worked example: travel time

A delivery truck holds a steady 60 km/h on a 150 km highway route. How long does the trip take?

Solve

for time:

That is 2 hours 30 minutes. Notice the kilometers cancel, leaving hours.

Average acceleration defined

Acceleration is the rate at which velocity changes. Average acceleration divides the change in velocity by the time the change took.

Units are meters per second per second, written m/s². A cart that gains 6 m/s in 2 s accelerates at 3 m/s².

Acceleration versus speed

Speed says how fast you are going; acceleration says how fast that velocity is changing. They are independent. A jet cruising at 250 m/s in a straight line has zero acceleration. A sprinter leaving the blocks has a low speed but a large acceleration. A large speed does not imply a large acceleration, and a small speed does not imply a small one.

Positive and negative acceleration

The sign of acceleration tells which way the velocity change points along your chosen axis. With right as positive, a positive acceleration means velocity is becoming more rightward; a negative one means it is becoming more leftward. The sign alone does not say whether the object speeds up or slows down. You must compare it with the sign of the velocity.

When an object slows down

An object slows down when its acceleration points opposite its velocity. A car moving in the positive direction with negative acceleration slows. A car moving in the negative direction with positive acceleration also slows. When velocity and acceleration share a sign, the object speeds up. "Deceleration" is just everyday language for this opposite-sign case.

v and a sign grid: same sign speeds up, opposite signs slow down.

Worked example: signed acceleration

Take east as positive. A cart rolls west at 6 m/s and, 2 s later, is rolling west at 2 m/s. Find its acceleration.

The acceleration is positive (east) while the cart moves west, so the cart is slowing down.

Cart moving west at −6 m/s then −2 m/s with acceleration +2 m/s² pointing east, so it slows down.

Constant acceleration as a model

Many test problems assume acceleration stays constant: a car pulling away steadily, a brake applied with even pressure, or an object falling with air resistance ignored. Real motion is rarely perfectly uniform, but this model is simple and close enough for short intervals. The kinematic equations in this lesson are only valid when acceleration is constant.

Velocity after a known time

With constant acceleration, velocity changes by the same amount each second. Start with the initial velocity and add the acceleration times the elapsed time.

Here is the starting velocity in m/s, is acceleration in m/s², and is time in s. Signs must follow your chosen positive direction.

Worked example: start from rest

A forklift starts from rest and accelerates evenly at 2.5 m/s² for 8 s. What is its speed at the end?

From rest means

Each second adds 2.5 m/s, and eight seconds of that adds 20 m/s.

Displacement during constant acceleration

When velocity changes steadily, displacement depends on both the starting velocity and the extra distance gained from acceleration.

The first term is how far you would go with no acceleration. The second term grows with the square of time, so doubling the time quadruples that part.

Worked example: acceleration from rest

A go-kart leaves the line from rest at 2 m/s² for 5 s. How far does it travel?

With

only the second term remains.

As a check, its final speed is 10 m/s, its average speed is 5 m/s, and 5 m/s times 5 s is 25 m.

Average velocity for constant acceleration

Under constant acceleration velocity changes in a straight line with time, so the average velocity is exactly halfway between the start and end values.

Multiply that average by the time to get displacement. This shortcut fails if acceleration is not constant.

Worked example: changing velocity evenly

A tram speeds up evenly from 4 m/s to 16 m/s over 6 s without reversing. Find its displacement.

Average velocity is

m/s

You did not need the acceleration (2 m/s²) at all; the endpoints and the time were enough.

Distance without elapsed time

Sometimes a problem gives speeds and acceleration but no time. Use the relation that skips time entirely.

For a stop, set

and use a negative when the motion is positive. Solving gives the braking distance

Units: (m/s)² divided by m/s² leaves meters.

Worked example: required stopping distance

A van moving at 25 m/s brakes steadily at 5 m/s² until it stops. How much road does braking need?

So the braking distance is 62.5 m. Ignoring reaction time, the van needs 62.5 m of clear road ahead.

Braking distance versus stopping distance

Braking distance is only the part covered after the brakes take hold. Stopping distance is the total: reaction distance plus braking distance. During the reaction time the driver has not yet pressed the pedal, so the vehicle keeps its full speed. Questions that say "total stopping distance" want both parts added together.

Reaction distance example

During reaction time the vehicle has no braking, so it moves at constant speed and

applies.

A driver going 22 m/s takes 1.0 s to react before braking. The reaction distance is

m

If the braking distance on that road is 30 m, the total stopping distance is

m
Stopping distance drawn to scale along a road: 22 m of reaction distance (22 m/s × 1.0 s) followed by 30 m of braking, total 22 + 30 = 52 m.

Effect of speed on braking distance

Because braking distance equals , it grows with the square of speed. With the same deceleration of 6.25 m/s²:

and 30 m/s needs 72 m. Doubling speed makes braking distance four times longer; tripling it makes it nine times longer. Reaction distance, by contrast, only doubles.

Braking distance at 6.25 m/s²: 8 m, 32 m (×4), 72 m (×9) for 10, 20, 30 m/s.

Motion graphs: position versus time

On a position-time graph, the slope equals velocity.

A rising straight line means constant positive velocity, a falling line means constant negative velocity, and a horizontal line means the object is not moving. A curved line means the velocity is changing.

Position-time graph with three straight lines from one point: rising means v > 0, flat means v = 0, falling means v < 0; a Δx over Δt triangle shows slope equals velocity.

Worked example: position graph slope

A position-time line starts at 10 m at t = 0 and falls in a straight line to 2 m at t = 4 s. Find the velocity.

Position-time line from 10 m to 2 m over 4 s: rise −8 m, run 4 s, v = ?

The object moves toward the negative direction at a steady 2 m/s.

Motion graphs: velocity versus time

On a velocity-time graph, two features carry meaning. The slope is acceleration: steep means rapid change, flat means constant velocity. The area between the line and the time axis is displacement. A line above the axis means motion in the positive direction; below the axis means motion in the negative direction.

Worked example: area under velocity graph

A scooter's velocity rises in a straight line from 0 to 8 m/s over 4 s, then holds at 8 m/s for 3 s. Find the displacement.

The triangle gives

m

The rectangle gives

m

Total displacement is

m
Scooter velocity-time graph: triangle area 16 m plus rectangle area 24 m gives 40 m.

Signed area and reversing direction

Area above the time axis counts as positive displacement; area below counts as negative. If a velocity line crosses the axis, the object turned around at that instant. To get displacement, add the signed areas. To get total distance, add the sizes of the areas and ignore the signs.

Worked graph: forward then backward

A cart's velocity graph shows +4 m/s for 3 s, then −2 m/s for 5 s.

First area:

m

Second area:

m

Displacement is

m

Distance is

m

The cart ends just 2 m ahead of its start after traveling 22 m.

Velocity-time graph: +12 m area above the axis and −10 m area below; displacement +2 m, distance 22 m.

Acceleration-time graph interpretation

On an acceleration-time graph, the area under the line equals the change in velocity.

A horizontal line at 1.5 m/s² lasting 4 s means velocity rises by 6 m/s. The graph tells you the change only; you still need the starting velocity to know the actual velocity.

a-t graph at 1.5 m/s² for 4 s: area = Δv = 6 m/s.

Uniform acceleration and graph shape

For constant, nonzero acceleration, the three graphs have signature shapes. The acceleration-time graph is a flat horizontal line. The velocity-time graph is a straight sloped line. The position-time graph is a curve (a parabola) that bends upward for positive acceleration and downward for negative. Recognizing these shapes answers many graph questions quickly.

For constant positive acceleration: flat a-t line, straight sloped v-t line, upward-curving x-t graph.

Free fall as accelerated motion

An object in free fall moves under gravity alone, with air resistance ignored. Near Earth its acceleration is about 9.8 m/s² straight down; tests often round to 10 m/s². Dropped from rest, its speed and drop distance after time t are

Each second adds about 9.8 m/s of downward speed.

Worked example: dropped object

A bolt slips from a scaffold and falls freely for 2.0 s. Use

g = 9.8 m/s²

and ignore air resistance.

Speed after 2.0 s:

Drop distance:

With g rounded to 10 m/s², both answers become 20.

Upward throw under gravity

A ball tossed straight up slows by about 9.8 m/s each second while rising, stops for an instant at the top, then speeds up on the way down. Throughout the flight, including at the top, the acceleration is g pointing downward. Only the velocity is zero at the peak. With no air drag, the trip up takes the same time as the trip down.

Worked upward throw

A ball leaves a hand straight upward at 15 m/s. Use

g = 10 m/s²
Ball thrown up at 15 m/s: velocity shrinks to 0 at the top and points down after; a = g = 10 m/s² down throughout; h = ?, time to top = ?

Time to the top, where velocity is zero:

s

It returns to the hand 3.0 s after release, moving at 15 m/s downward.

Mass independence in ideal free fall

When air resistance is negligible, every object falls with the same acceleration regardless of its mass. A heavier hammer feels a bigger gravitational pull, but it also has more mass to accelerate, and the two effects cancel. Dropped together in a vacuum, a hammer and a feather land at the same time. In air, light or spread-out objects fall slower because of drag.

Circular motion and direction change

Velocity includes direction, so an object moving in a circle at constant speed is still accelerating. That acceleration points toward the center of the circle and is called centripetal acceleration.

Here v is speed in m/s and r is the radius in m, giving m/s².

Circular motion: velocity along the path, acceleration a = v²/r toward the center.

Worked turn-radius comparison

A cart takes a curve at 10 m/s. Compare a 50 m radius with a 25 m radius.

Halving the radius doubles the acceleration. Doubling the speed on the same curve would quadruple it, because speed is squared.

Relative motion in one dimension

Velocity always depends on who is watching. The velocity of A as seen by B is A's velocity minus B's velocity, both measured from the ground.

A passenger walking forward at 1 m/s on a train moving at 20 m/s moves at 21 m/s relative to the ground.

Train 20 m/s plus walker 1 m/s gives 21 m/s relative to the ground.

Head-on closing speed

When two objects move toward each other, the gap shrinks at the sum of their speeds.

Two carts 90 m apart roll toward each other at 9 m/s and 6 m/s. They close at 15 m/s and meet after

s
Carts 90 m apart at 9 m/s and 6 m/s toward each other: closing speed 15 m/s.

Same-direction closing speed

When a faster object chases a slower one in the same direction, the gap shrinks at the difference of their speeds.

A rover at 14 m/s trails a rover at 10 m/s by 32 m. The gap shrinks by 4 m each second, so it catches up after

s
Rover at 14 m/s 32 m behind one at 10 m/s: closing speed 4 m/s.

Two-stage journeys

Many problems split motion into stages, such as speeding up and then cruising. Treat each stage as its own problem. The final velocity of stage one becomes the initial velocity of stage two. Find the displacement of each stage with the formula that fits it, then add them. Never apply one formula across a boundary where the acceleration changes.

Worked example: acceleration then cruise

A shuttle cart starts from rest, accelerates at 2 m/s² for 6 s, then cruises for 10 s.

v-t graph: accelerate at 2 m/s² for 6 s, then cruise 10 s; v and both stage distances = ?

Stage one:

m/s
and
m

Stage two:

m

Total distance is

m

Average speed over unequal periods

Average speed is total distance over total time, not the average of the speeds.

A bike rides 3 m/s for 10 s (30 m) and 9 m/s for 5 s (45 m). Average speed is

m/s

not 6 m/s, because more time was spent going slowly.

When average acceleration is zero

Average acceleration is zero whenever the final velocity equals the initial velocity, even if the object sped up and slowed down in between. A bus that leaves a stop, travels, and halts at the next stop has zero average acceleration for that whole leg. A belt moving at one steady velocity also has zero acceleration at every moment.

Rounding and significant units

Keep full precision during a calculation and round only at the end, or small errors can grow. Match the precision of the given data: if values have two significant figures, report about two. Always attach units, because a number like 20 could mean m, m/s or m/s². On multiple-choice items, an answer with the wrong unit is wrong.

Dimensional analysis of motion formulas

Checking units catches many errors. Each term in a valid equation must have the same unit. In the displacement formula,

and also gives meters. If an expression you build gives m/s when you need meters, a time factor is missing.

Choosing the right equation

List the known quantities and the unknown, then pick the equation that contains exactly those. No distance asked: use

No final velocity given: use

No time given: use

No acceleration given: use average velocity times time.

Exam trap: zero speed at one instant

An object can have zero velocity for an instant while still accelerating. At the top of a toss, or when a cart reverses, velocity passes through zero, but gravity or the pushing force is still changing it. Zero velocity does not mean zero acceleration, and zero acceleration does not require zero velocity.

Exam trap: negative means braking

A negative acceleration does not automatically mean an object is slowing down. It only means the acceleration points in the negative direction. If the object is already moving in the negative direction, a negative acceleration speeds it up. Check whether velocity and acceleration have the same sign (speeding up) or opposite signs (slowing down).

Exam trap: mixing distance and graph area

The signed area under a velocity-time graph gives displacement, not distance. If part of the graph lies below the time axis, signed areas cancel and the net result can be small or even zero. When a question asks for total distance traveled, add the area sizes without signs. Read the question carefully to see which one is wanted.

Exam trap: doubling acceleration or speed

Scaling effects are not all linear. For a fixed time, doubling acceleration doubles the distance gained from rest. For a fixed acceleration, doubling time quadruples that distance. For braking with fixed deceleration, doubling initial speed quadruples braking distance but only doubles the stopping time. Know which quantity is squared.

Guided drill: three quick decisions

Is a car circling a track at a constant 15 m/s accelerating? Answer: yes, because its direction keeps changing.

A cart moves at −3 m/s with +1 m/s² acceleration. Speeding up or slowing down? Answer: slowing, because the signs differ.

A v-t graph line is flat at 7 m/s. What is the acceleration? Answer: zero.

Worked example: speed is not velocity

A runner completes one 400 m lap of a track in 80 s and finishes at the starting line.

Because she ends where she began, her displacement is zero, so her average velocity is 0 m/s. Same trip, same time, but the two averages differ because one uses path length and the other uses net change in position.

Reading a motion table

Tables of data can reveal acceleration without a graph. A test cart's velocity is recorded each second: 2, 5, 8 and 11 m/s at t = 0, 1, 2 and 3 s.

The velocity rises by 3 m/s every second, so acceleration is a constant 3 m/s².

Equal gains in equal times signal constant acceleration.

Instantaneous acceleration from a curve

If a velocity-time graph is curved, acceleration changes from moment to moment. To find it at one instant, imagine a straight line that just touches the curve at that point, the tangent line. Its slope is the instantaneous acceleration. Where the curve is steep, acceleration is large; where it flattens out, acceleration approaches zero.

Curved v-t graph: steep tangent = large a, flat tangent = a ≈ 0.

Reverse direction under constant acceleration

A puck moves at +6 m/s up a ramp and has a constant acceleration of −2 m/s². It stops after

s

having moved 9 m. It then slides back. At t = 5 s:

Distance covered is 9 m up plus 4 m back, or 13 m.

Puck v-t line from +6 to −4 m/s, turning at 3 s: +9 m and −4 m areas; displacement 5 m, distance 13 m.

Displacement is not always distance

Displacement is a straight-line arrow, while distance follows the actual path. A surveyor walks 30 m east, then 40 m north. Distance is 70 m. Displacement is the diagonal from start to finish:

30 m east then 40 m north: distance 70 m, diagonal displacement = ?

Displacement equals distance only for straight motion in one direction without reversing. Any turn or backtrack makes distance larger.

Deriving the stopping-time relation

Start from

At a stop the final velocity is zero, and the acceleration has size opposing the motion, so

A train at 30 m/s braking at 6 m/s² stops in

s

Deriving distance using average speed

During a steady stop, speed falls evenly from to zero, so the average speed is . Multiply by the stopping time:

Example: a train moving at m/s brakes at m/s² and stops in s, so

m

The time-free formula agrees:

m

Two routes, one answer.

Why curved-path acceleration matters

Turning requires a force toward the center of the curve. For a car, that inward push comes from friction between tires and road. Since the needed acceleration is , a faster turn or a tighter curve demands much more grip. On ice or loose gravel the available friction may be too small, and the vehicle slides outward along a straighter path. That is why curves carry lower posted speeds.

Acceleration and mechanical causes

Acceleration does not happen on its own; a net force causes it. Newton's second law links the two:

A 1200 kg car with 3000 N of net forward force accelerates at

m/s²

The same force on a heavier vehicle produces less acceleration.

Ideal assumptions and real workshop machines

Textbook kinematics assumes point-like objects, constant acceleration, no air drag, and instant changes between stages. Real machines differ: motors ramp up gradually, belts stretch, brakes fade as they heat, and air resistance grows with speed. Ideal answers are best treated as estimates. When safety depends on a stopping or clearance distance, engineers add margin beyond the ideal figure.

Graph intersection and pursuit

When two objects are plotted on the same position-time graph, the point where their lines cross is the moment they are at the same place, so it marks a meeting or a pass. On a velocity-time graph, a crossing means only that they share the same velocity at that instant, not that they are side by side. Check which graph you are reading before you answer a pursuit question.

Two position-time lines crossing: same position, meet or pass.

Acceleration on different coordinate axes

You may choose any axis direction, as long as signs follow it. For vertical motion with up positive, gravity is

m/s²

with down positive it is +9.8 m/s². For a cart on a ramp, pick an axis along the slope, so motion and acceleration lie on one line. The physics does not change; only the signs on the numbers change.

Speed after distance with no stopwatch

If you know the acceleration and the distance but not the time, use the time-free relation to find speed.

A cart starts from rest and accelerates at 3 m/s² over 24 m. Then

so
m/s

No time measurement was needed.

Final integrated self-check

A cart at 6 m/s brakes at 2 m/s². How long to stop? Answer: 3 s.

How far does it go while stopping? Answer:

m

A ball is tossed up at 20 m/s with

g = 10 m/s²

What is its acceleration at the top? Answer: 10 m/s² downward.

Convert 36 km/h to m/s. Answer: 10 m/s.

A runner laps a track and ends at the start. Average velocity? Answer: zero.

Review checklist and next connection

Before moving on, confirm you can: set an origin and a positive direction; separate distance from displacement and speed from velocity; apply the constant-acceleration equations with signs; read slope and area on motion graphs; solve free-fall, braking and two-stage problems; and check units. The next lessons connect these motion ideas to forces, friction and Newton's laws.

Quick review

Distance vs. displacement: what is the difference?
Distance is the total path length traveled (never negative). Displacement is the net change in position, , with a sign or direction. A round trip has zero displacement but nonzero distance.
Constant acceleration: how do you find velocity after time t?
, where is starting velocity (m/s), a is acceleration (m/s²) and t is time (s). Example: from rest at 2.5 m/s² for 8 s gives 20 m/s.
Constant acceleration: displacement after time t?
(meters). The first term is travel with no acceleration; the second grows with time squared, so doubling t quadruples that part.
Which equation links speeds, acceleration and distance with no time given?
. Setting v = 0 gives braking distance , so doubling speed makes braking distance 4 times longer.
When does an object speed up or slow down under acceleration?
Same sign for velocity and acceleration: speeding up. Opposite signs: slowing down. Negative acceleration alone does not mean braking; an object moving in the negative direction speeds up with it.
Motion graphs: what do slope and area tell you?
Position-time: slope = velocity. Velocity-time: slope = acceleration, area under the line = displacement (area below the axis counts negative; add area sizes for total distance).
Free fall: what is the acceleration, and what is it at the top of an upward throw?
About 9.8 m/s² (often 10) downward for every mass if air drag is ignored. At the top of a throw velocity is zero but acceleration is still g downward. Dropped from rest: , .
Constant speed around a circle: is it accelerating, and how much?
Yes, because direction changes. Centripetal acceleration points to the center: (v in m/s, r in m). Halving the radius doubles it; doubling speed quadruples it.
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About this lesson

Constant-acceleration sections cover velocity after a known time, displacement from rest, stopping distance with and without reaction time, and how speed affects braking distance. You will read position-time, velocity-time and acceleration-time graphs, including signed area when an object reverses. Free fall, upward throws, circular motion, relative motion with closing speeds, and two-stage journeys round out the content, followed by exam traps and a guided drill.

On the ASVAB, Mechanical Comprehension can present motion as a quick graph or table reading or as a short calculation, and this lesson prepares you for both. It is one of the longer ones at about 120 minutes. Close it out with the 40-question lesson test.

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