Motion and Acceleration flashcards

Distance is the total path length traveled (never negative). Displacement is the net change in position, $\Delta x=x_f-x_i$, with a sign or direction. A round trip has zero displacement but nonzero distance.

$v=v_0+at$, where $v_0$ 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.

$\Delta x=v_{0}t+\frac{1}{2}a{t}^2$ (meters). The first term is travel with no acceleration; the second grows with time squared, so doubling t quadruples that part.

$v^2=v_0^2+2a\Delta x$. Setting v = 0 gives braking distance $\Delta x=\frac{v_0^2}{2|a|}$, so doubling speed makes braking distance 4 times longer.

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.

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).

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: $v=gt$, $d=\frac{1}{2}g{t}^2$.

Yes, because direction changes. Centripetal acceleration points to the center: $a_c=\frac{v^2}{r}$ (v in m/s, r in m). Halving the radius doubles it; doubling speed quadruples it.