Motion and Acceleration: Lesson Test

Answer: B

Correct answer: 35 m. Distance is the total path length, so the two legs simply add: 25 m + 10 m = 35 m. Direction does not matter for distance.

• A: This counts only the first leg and leaves out the 10 m walked back.
• B (correct): Distance adds the path lengths without signs: 25 m plus 10 m is 35 m.
• C: 15 m east is the displacement, the net change in position, not the distance walked.
• D: Distance is never negative, and 15 m is the size of the displacement, not the path length.

Answer: C

Correct answer: Her displacement. Displacement is the change in position, final minus initial. Because she ends at her starting point, that change is zero even though she covered a real distance at a nonzero average speed.

• A: Distance is the full path length, which grew with every step of the loop.
• B: Average speed uses the distance walked, which is greater than zero.
• C (correct): Displacement is final position minus initial position, and both are the same point.
• D: The walk took real time; only the net change in position vanishes.

Answer: B

Correct answer: 15 m/s. To convert km/h to m/s, divide by 3.6.

• A: This keeps the number unchanged and only swaps the unit label.
• B (correct): Dividing km/h by 3.6 gives m/s: 54 divided by 3.6 equals 15.
• C: This divides by 60 as if converting minutes, not converting km/h to m/s.
• D: This multiplies by 3.6, which is the step for going from m/s to km/h.

Answer: D

Correct answer: -12 m. With north positive, any point south of the origin has a negative position. The bench is 12 m south, so its position is -12 m.

• A: This ignores the sign convention; south is the negative direction here.
• B: Zero would place the bench at the bait shop itself, which is the origin.
• C: This doubles the distance and also uses the wrong sign.
• D (correct): The bench is 12 m from the origin in the negative (south) direction.

Answer: D

Correct answer: A straight line sloping at a steady angle. For constant acceleration, the slope of the velocity-time graph equals that acceleration and does not change. A constant slope is a straight sloped line.

• A: A line on the axis means zero velocity, so the car would not be moving at all.
• B: The parabola shape belongs to the position-time graph, not the velocity-time graph.
• C: A flat velocity line means constant velocity, which is zero acceleration.
• D (correct): Constant acceleration means velocity changes by the same amount each second, giving a straight, sloped line.

Answer: A

Correct answer: 3 m/s². Average acceleration is change in velocity over time:

• A (correct): The velocity changes by 12 m/s in 4 s, and 12 divided by 4 is 3 m/s².
• B: This divides the time by the change in velocity, which is upside down.
• C: This divides the final speed of 14 m/s by the time and ignores the starting speed of 2 m/s.
• D: This multiplies the change in velocity by the time instead of dividing.

Answer: B

Correct answer: 4 m/s. Average speed equals total distance divided by total time.

• A: This multiplies distance by time instead of dividing.
• B (correct): Average speed is distance over time: 1,200 m divided by 300 s equals 4 m/s.
• C: This misplaces the decimal; 1,200 divided by 300 is 4, not 40.
• D: This divides time by distance instead of distance by time.

Answer: D

Correct answer: 0 m/s². Acceleration measures how quickly velocity changes. Constant speed in a straight line means the velocity never changes, so the acceleration is zero no matter how fast the plane is going.

• A: This confuses speed with acceleration; a high speed does not mean the velocity is changing.
• B: That is the acceleration of free fall, but a plane in steady level flight is not falling.
• C: There is no change in velocity to divide by time, so no nonzero acceleration applies.
• D (correct): Its speed and direction are both constant, so its velocity is not changing and its acceleration is zero.

Answer: D

Correct answer: They reach the floor at the same time. In ideal free fall the acceleration is g for every object. A heavier object feels a larger gravitational force but has proportionally more mass to accelerate, so both balls land together.

• A: The larger pull on the heavier ball is matched by its larger mass, so it gains no head start.
• B: Gravity still acts in a vacuum; only air drag has been removed.
• C: Lighter objects do not fall faster; mass does not change free-fall acceleration.
• D (correct): With no air resistance, every object falls with the same acceleration regardless of mass.

Answer: B

Correct answer: Yes, because its direction keeps changing. Acceleration is any change in velocity, and velocity includes direction. Moving in a circle constantly changes direction, so the go-kart has a centripetal acceleration pointing toward the center of the track.

• A: The speed is stated as steady; the acceleration comes from the turning, not from speeding up.
• B (correct): Velocity includes direction, so a changing direction means changing velocity and therefore acceleration.
• C: Returning to the start makes the lap displacement zero but says nothing about acceleration during the turn.
• D: Constant speed is not constant velocity when the direction keeps changing.

Answer: C

Correct answer: 1 m/s east. Displacement is

m and total time is 40 s + 20 s = 60 s.

The positive sign means 1 m/s east.

• A: This uses only the first leg, 80 m in 40 s, and ignores the walk back.
• B: This divides the 60 m displacement by only the 40 s of the first leg.
• C (correct): Displacement is 80 m minus 20 m, or +60 m, and 60 m divided by 60 s is 1 m/s east.
• D: This uses total distance (100 m) over 60 s, which is average speed, not average velocity.

Answer: A

Correct answer: 39 m. Displacement is the area under the velocity-time graph.

Triangle:

Rectangle:

Total:

• A (correct): The triangle gives one-half of 3 times 6, or 9 m, and the rectangle gives 6 times 5, or 30 m, for 39 m total.
• B: This averages 0 and 6 m/s and applies 3 m/s to all 8 s, using the speed-up shortcut across a stage boundary.
• C: This counts only the flat 5 s section and leaves out the triangle.
• D: This treats the robot as moving at 6 m/s for all 8 s, ignoring the speed-up stage.

Answer: C

Correct answer: 45 m. For a drop from rest, use

• A: This is g times t, which gives the speed in m/s after 3 s, not the distance.
• B: This uses g t squared and leaves out the one-half factor.
• C (correct): Drop distance is one-half g t squared: 0.5 times 10 times 9 equals 45 m.
• D: This multiplies one-half by g by t without squaring the time.

Answer: C

Correct answer: 18 m/s. Use

• A: This assumes the speed stays the same, but a steady acceleration changes it every second.
• B: 12 m/s is only the speed gained; the truck's starting speed of 6 m/s must be added.
• C (correct): Using v = v0 + at gives 6 + 1.5 times 8, which is 6 + 12 = 18 m/s.
• D: This adds the acceleration to the speed only once instead of once for each of the 8 seconds.

Answer: D

Correct answer: 24 m. From rest,

so only the acceleration term remains:

• A: This leaves out the one-half in the formula, doubling the true distance.
• B: This uses time instead of time squared, giving one-half times 3 times 4.
• C: 12 is the final speed in m/s (3 times 4), not the distance traveled.
• D (correct): From rest, distance is one-half times 3 times 4 squared, which is 0.5 times 3 times 16 = 24 m.

Answer: B

Correct answer: 50 m. With constant acceleration, average velocity is halfway between the endpoints:

Then

• A: This uses only the starting speed and ignores the speeding up.
• B (correct): With constant acceleration the average speed is (6 + 14)/2 = 10 m/s, and 10 times 5 s is 50 m.
• C: This multiplies the change in speed (8 m/s) by the time instead of using the average speed.
• D: This uses the final speed for the whole interval, but the bus was slower at the start.

Answer: C

Correct answer: 7.2 m. At the top the velocity is zero, so

• A: This divides 144 by 2 and forgets to divide by g.
• B: This divides 144 by 10 and leaves out the factor of 2, the same as using the full launch speed for the whole climb.
• C (correct): Maximum height is v0 squared over 2g: 144 divided by 20 equals 7.2 m.
• D: This is the time in seconds to reach the top, not the height in meters.

Answer: C

Correct answer: It is speeding up, because velocity and acceleration have the same sign. Moving west means the velocity is negative. The acceleration is also negative, so velocity and acceleration point the same way and the cart speeds up. A negative acceleration means slowing only when the velocity is positive.

• A: Signs do not cancel this way; any nonzero acceleration changes the velocity.
• B: A negative sign only gives direction; braking happens when acceleration opposes velocity.
• C (correct): Moving west gives a negative velocity, and with a negative acceleration both point west, so speed increases.
• D: The direction of motion alone does not decide it; you must compare velocity and acceleration signs.

Answer: B

Correct answer: 10 s. When objects approach head-on, the closing speed is the sum of their speeds:

• A: This uses only the faster cart's speed, giving about 17 s.
• B (correct): Moving head-on, the gap closes at 7 + 5 = 12 m/s, and 120 m divided by 12 m/s is 10 s.
• C: This uses the difference of the speeds, which applies only when one object chases another.
• D: This uses only the slower cart's speed and ignores the other cart's motion.

Answer: A

Correct answer: 80 m. Braking distance equals $v_0^2/(2|a|)$, so it scales with the square of speed. Going from 15 m/s to 30 m/s doubles the speed, which makes the braking distance

times larger: 4 times 20 m = 80 m.

• A (correct): Doubling speed multiplies braking distance by 2 squared, or 4, so 20 m becomes 80 m.
• B: This multiplies by 2 cubed; braking distance depends on the square of speed, not the cube.
• C: This triples the 20 m, but doubling the speed multiplies braking distance by 2 squared, not by 3.
• D: This assumes braking distance grows in direct proportion to speed, but it grows with the square.

Answer: B

Correct answer: 50 m. Use the time-free relation for a stop:

• A: 20 divided by 4 is the stopping time in seconds, not a distance.
• B (correct): Braking distance is 20 squared divided by 2 times 4, which is 400 divided by 8 = 50 m.
• C: This divides the speed, not the speed squared, by twice the deceleration.
• D: This divides by the deceleration alone and leaves out the factor of 2 in the denominator.

Answer: C

Correct answer: 3 m/s. Velocity is the slope of a position-time graph:

• A: This divides time by displacement, which turns the slope upside down.
• B: This divides the final position, 22 m, by the time instead of using the change in position.
• C (correct): The slope is (22 - 4) divided by (6 - 0), which is 18 divided by 6 = 3 m/s.
• D: 18 m is the displacement; it still has to be divided by the 6 s elapsed.

Answer: C

Correct answer: +2 m/s². With north positive, the velocities are -10 m/s and -2 m/s.

The acceleration points north while the car moves south, so the car is slowing down.

• A: This adds 10 and 2 to get 12 m/s; the change from -10 m/s to -2 m/s is only 8 m/s.
• B: This adds the two speeds instead of subtracting signed velocities, and also gives the wrong sign.
• C (correct): Final minus initial velocity is (-2) - (-10) = +8 m/s, and 8 divided by 4 s is +2 m/s².
• D: The car is moving south but slowing, so its acceleration points north, which is the positive direction.

Answer: B

Correct answer: 3.6 s. On the way up the ball loses 10 m/s each second, so it stops after

With no air drag the trip down takes the same time, so the total is

• A: This doubles the correct flight time; the ball takes 1.8 s up and 1.8 s down.
• B (correct): The ball takes 18 divided by 10 = 1.8 s to stop at the top and the same time to fall back, 3.6 s in all.
• C: 1.8 s is only the time to reach the top; the trip back down takes just as long.
• D: This divides the launch speed by 2g, which is half the time to reach the top.

Answer: D

Correct answer: 23.5 m/s forward. Take forward as positive. The train moves at +25 m/s, and the passenger moves at -1.5 m/s relative to the train. Relative to the ground: 25 + (-1.5) = +23.5 m/s, which is 23.5 m/s forward.

• A: That is the velocity relative to the train, not relative to the ground.
• B: This ignores the passenger's own walking, which does change the velocity seen from the ground.
• C: This adds the walking speed as if the passenger were walking toward the front of the train.
• D (correct): Walking backward subtracts from the train's motion: 25 - 1.5 = 23.5 m/s, still forward.

Answer: A

Correct answer: 3 m/s². Newton's second law links force and acceleration:

• A (correct): Acceleration is net force divided by mass: 4,500 divided by 1,500 is 3 m/s².
• B: This divides the mass by the force, which turns Newton's second law upside down.
• C: That is the free-fall acceleration from gravity, not the result of a 4,500 N push on this tractor.
• D: This slips a decimal place; 4,500 divided by 1,500 is 3, not 30.

Answer: B

Correct answer: The carts have equal velocities at t = 6 s. A velocity-time graph plots velocity against time, so where two lines meet the carts share the same velocity at that instant. It says nothing about whether they are at the same position; that would need a position-time graph.

• A: Overtaking requires equal positions, which a velocity-time crossing does not show.
• B (correct): Each point on a velocity-time graph gives velocity, so a shared point means the same velocity at that moment.
• C: Acceleration is the slope of each line, and the lines can cross while both still have slope.
• D: Being at the same place is shown by a crossing on a position-time graph, not a velocity-time graph.

Answer: D

Correct answer: 4 m/s². Use

• A: This doubles the correct value, as if borrowing the 2 from a braking formula.
• B: This divides the radius by the speed, which is upside down and unsquared.
• C: This divides speed by radius without squaring the speed.
• D (correct): Centripetal acceleration is v squared over r: 144 divided by 36 equals 4 m/s².

Answer: A

Correct answer: 9 s. In a same-direction chase, the gap closes at the difference of the speeds:

• A (correct): The gap shrinks at 16 - 11 = 5 m/s, and 45 m divided by 5 m/s is 9 s.
• B: This is the closing speed in m/s mistaken for the catch-up time.
• C: This divides the gap by the patrol vehicle's full speed and ignores the truck's motion.
• D: This adds the speeds, which is correct only for objects moving toward each other.

Answer: A

Correct answer: 13 m/s. The area under an acceleration-time graph is the change in velocity:

Add it to the starting velocity:

• A (correct): The area 2.5 times 4 gives a 10 m/s gain, and 3 + 10 = 13 m/s.
• B: This is only the change in velocity from the graph area; the starting 3 m/s must be added.
• C: This uses one-half a t squared, which gives a distance from rest, not a velocity.
• D: This adds the acceleration value directly to the velocity without multiplying by time.

Answer: D

Correct answer: 2 m/s. Average velocity uses displacement. The two legs form a right angle, so

Average velocity has size

• A: This subtracts the two legs to get 20 m, but the legs are at right angles, so they combine by the Pythagorean rule.
• B: This uses the total distance of 140 m, which gives average speed, not average velocity.
• C: This averages the two leg lengths to 70 m instead of finding the straight-line displacement.
• D (correct): Her displacement is the square root of 60 squared plus 80 squared, which is 100 m, and 100 m divided by 50 s is 2 m/s.

Answer: D

Correct answer: 192 m. Stage one:

Stage two:

Total:

• A: This assumes the cart moved at its top speed of 12 m/s for the entire 20 s.
• B: This applies the acceleration formula over all 20 s, but the cart stopped accelerating after 8 s.
• C: This counts only the cruising stage and leaves out the 48 m covered while speeding up.
• D (correct): Stage one covers 48 m and reaches 12 m/s; stage two covers 12 times 12 = 144 m; 48 + 144 = 192 m.

Answer: A

Correct answer: 20 m. Stop time:

Distance up:

From 4 s to 6 s it moves back down from rest:

Total distance is

even though the displacement is only +12 m.

• A (correct): The puck stops at 4 s after 16 m up, then moves 4 m back in the last 2 s, for 16 + 4 = 20 m of path.
• B: This is the displacement, 48 - 36 = 12 m, which subtracts the return trip instead of adding it.
• C: This counts only the trip up the ramp and leaves out the return trip down.
• D: This adds the acceleration term instead of subtracting it, using +36 m where -36 m belongs.

Answer: D

Correct answer: 6 s and 72 m. The deceleration is

From 24 m/s:

Doubling the speed doubles the stopping time but makes the braking distance four times the original 18 m.

• A: This quadruples the time and only doubles the distance, which reverses which quantity is squared.
• B: Stopping time is speed divided by deceleration, so doubling the speed only doubles the time to 6 s.
• C: The time is right, but braking distance grows with the square of speed, so it becomes four times 18 m, not twice.
• D (correct): The deceleration is 4 m/s², so the time is 24 divided by 4 = 6 s and the distance is 24 squared divided by 8 = 72 m.

Answer: B

Correct answer: 7.5 m/s. Find each distance:

m and

• A: This averages the two speeds, which ignores that far more time was spent at the slower speed.
• B (correct): Each stage covers 300 m, so 600 m divided by 80 s gives 7.5 m/s.
• C: This divides only one stage's 300 m by the total 80 s.
• D: This divides the total 600 m by only the first stage's 20 s.

Answer: A

Correct answer: 54 m. Reaction distance at constant speed:

Braking distance:

Total stopping distance:

• A (correct): Reaction distance is 24 times 0.75, or 18 m, and braking distance is 576 over 16, or 36 m, for 54 m total.
• B: This is only the reaction distance, covered before the brakes are applied.
• C: This drops the 2 in the braking formula, giving 72 m of braking instead of 36 m.
• D: This is the braking distance alone and leaves out the 18 m covered during the reaction time.

Answer: C

Correct answer: 38 m. Area above the axis:

Area below the axis:

m in size.

Distance ignores signs, so it is

The displacement would be

• A: This adds the two speed values and treats the sum as a distance.
• B: This counts only the area above the axis and ignores the backward stage.
• C (correct): Add the area sizes without signs: 5 times 4 is 20 m and 3 times 6 is 18 m, for 38 m.
• D: This adds the signed areas, which gives the displacement, not the distance.

Answer: C

Correct answer: 40.5 m. The velocity drops 4 m/s every second, so the acceleration is -4 m/s². The stopping distance from 18 m/s is

Check: it stops after 4.5 s at an average of 9 m/s, and 9 times 4.5 is 40.5 m.

• A: This assumes the cart stops at t = 4 s, but at 4 m/s² it still has 2 m/s then and stops at 4.5 s.
• B: This leaves out the factor of 2 in the braking-distance formula, doubling the answer.
• C (correct): The cart slows by 4 m/s each second, so it travels 18 squared divided by 2 times 4 = 324/8 = 40.5 m.
• D: 4.5 is the time in seconds the cart takes to stop, not the distance it covers.

Answer: A

Correct answer: -10 m/s, 15 m above. With up positive,

The ball peaked at 2 s and is now moving down, 15 m above the hand.

• A (correct): Velocity is 20 - 10 times 3 = -10 m/s, and position is 20 times 3 minus 5 times 9 = 15 m above release.
• B: This adds the gravity term instead of subtracting it; with up positive, the acceleration is -10 m/s².
• C: The position is right, but after 3 s the ball is already falling, so its velocity is negative.
• D: 25 m is the total distance traveled (20 m up and 5 m back down), not the ball's position.

Answer: A

Correct answer: 67.5 m. Stage one:

Stage two:

Total:

• A (correct): The cart reaches 15 m/s after 45 m, then brakes over 225 divided by 10, or 22.5 m, for 67.5 m total.
• B: This leaves the 2 out of the braking formula, making the braking part 45 m instead of 22.5 m.
• C: This counts only the speed-up stage and leaves out the braking stage.
• D: This is only the braking stage; the 45 m speed-up stage must be added.

Answer: A

Correct answer: 130 m. Displacement is the straight line from start to finish. The legs are perpendicular, so use the Pythagorean theorem.

• A (correct): The two legs are at right angles, so the displacement is √(50² + 120²) = √16,900 = 130 m.
• B: This is the total distance flown along the path, not the straight-line displacement.
• C: This subtracts the legs as if they lay along one line in opposite directions.
• D: This averages the two legs, which has no meaning for displacement.

Answer: C

Correct answer: 300 m. Match the units first: 90 km/

Then d = vt =

• A: This multiplies 90 by 12 without converting km/h to m/s, mixing hours and seconds.
• B: This multiplies by 3.6 to convert, but going from km/h to m/s requires dividing by 3.6.
• C (correct): Converting 90 km/h gives 25 m/s, and 25 × 12 = 300 m.
• D: This slips a decimal place after the correct conversion to 25 m/s.

Answer: A

Correct answer: It is at rest. On a position-time graph, slope equals velocity. A horizontal line has zero slope, so the velocity is zero and the forklift stays put 8 m from the origin.

• A (correct): The slope of a position-time graph is velocity, and a horizontal line has zero slope.
• B: The 8 m is a position, not a speed; a flat line means the position never changes.
• C: Steady acceleration makes a curved position-time graph, not a flat line.
• D: Moving toward the origin would make the line slope downward, not stay flat.

Answer: B

Correct answer: 12 m/s. With no time given, use the time-free relation.

So

• A: This stops at v² = 144 and forgets to take the square root.
• B (correct): From v² = 0 + 2(4)(18) = 144, the speed is √144 = 12 m/s.
• C: This leaves out the factor of 2, computing √(4 × 18) = √72.
• D: This simply multiplies the acceleration by the distance, which does not give a speed.

Answer: A

Correct answer: −2 m/s². Write both velocities with signs:

The acceleration points backward, the negative direction.

• A (correct): The velocity changes from +4 to −6 m/s, a change of −10 m/s, and −10 ÷ 5 = −2 m/s².
• B: This uses only the difference in speeds, 6 − 4, and ignores that the direction reversed.
• C: The size is right, but the velocity change points backward, so the sign must be negative.
• D: This ignores both the reversal and the direction of the change.

Answer: B

Correct answer: 3 s. The ground is 15 m below the release point, so set the displacement to −15 m with

Rearranged:

which factors as

The positive root is t = 3 s.

• A: This is only the time to reach the top of the toss, where the velocity is momentarily zero.
• B (correct): Solving −15 = 10t − 5t² gives t² − 2t − 3 = 0, so t = 3 s.
• C: This is the time to rise and fall back to roof level; the hammer still has 15 m left to fall.
• D: This treats the hammer as simply dropped from 15 m and ignores the upward toss.

Answer: C

Correct answer: 16 s and 320 m. They meet when their positions match: car

patrol

Setting

Distance:

m (check:

).

• A: At 8 s the two speeds are equal, but the car is still ahead; equal velocity is not the same as meeting.
• B: This is where the patrol vehicle is when the speeds match, while the car is already at 160 m.
• C (correct): Setting 20t = 1.25t² gives t = 16 s, and both vehicles are then 20 × 16 = 320 m from the start.
• D: The time is right, but 640 m multiplies the final speed of 40 m/s by the time instead of using the average speed.

Answer: B

Correct answer: 36 m. Displacement is the area under a velocity-time graph. Here the shape is a triangle with base 6 s and height 12 m/s.

The same result comes from average speed:

• A: This uses the full rectangle, 12 × 6, as if the cart kept its starting speed the whole time.
• B (correct): The area under the line is a triangle: ½ × 6 × 12 = 36 m.
• C: This finds the slope, 12 ÷ 6, which is the size of the deceleration in m/s², not a distance.
• D: This takes half of the triangle's area by halving twice.

Answer: D

Correct answer: About 30.6 m. For a drop from rest,

• A: This is the speed reached after 2.5 s, 9.8 × 2.5, in m/s, not the distance fallen.
• B: This leaves out the one-half in d = ½gt².
• C: This forgets to square the time, computing ½ × 9.8 × 2.5.
• D (correct): The drop is ½ × 9.8 × 2.5² = 4.9 × 6.25 ≈ 30.6 m.

Answer: D

Correct answer: Instantaneous speed. Instantaneous speed is how fast an object is moving at a single moment. A speedometer reads exactly that. It shows no direction, so it is not a velocity, and it does not average over the trip.

• A: Average speed needs total distance and total time; the needle shows only the current moment.
• B: Average velocity uses displacement over elapsed time and includes direction, which a speedometer does not show.
• C: Acceleration is the rate of change of velocity; the speedometer shows the speed itself.
• D (correct): A speedometer shows how fast the truck is going at that moment, with no direction.