GRE Physics Companion: Acceleration

This entry is the GRE-oriented companion to M01-03. The core article develops average and instantaneous acceleration, graph interpretation, sign conventions, vector acceleration, circular motion, and reference-frame behavior. Here the emphasis is rapid classification and test-day efficiency.

1 Fast test-day strategy

When an acceleration problem appears, identify the structure first:

Average acceleration:
$a = Δv--. avg Δt$ (1)

Use the change in velocity, including direction.
Instantaneous acceleration:
$dv d2r a = dt-= dt2.$ (2)

Differentiate velocity once or position twice.
Velocity-time graph:
$ax = slope of the vx− t graph.$ (3)
One-dimensional speeding/slowing: compare the signs of $vx$ and $ax$ .
Circular motion at constant speed:
$v2 ac = --- R$ (4)

and the direction is inward.

Figure 1. GRE-speed acceleration triage: velocity change, derivatives, graph slope, sign comparison, and inward centripetal acceleration.

2 High-frequency traps

Negative acceleration does not automatically mean slowing down.
At the top of a vertical toss, velocity is zero but acceleration is not.
Constant speed does not imply zero acceleration if direction changes.
On a velocity-time graph, acceleration is slope, not height.
In a Galilean inertial-frame change, acceleration is unchanged.
In free fall without air resistance, different masses have the same gravitational acceleration at the same location.

Figure 2. Two frequent GRE traps: sign of acceleration versus change of speed, and tangent slope on a velocity-time graph.

3 Worked GRE example 1: differentiate efficiently

A particle has position

x(t) = 2t3 − 9t2 + 12t.

(5)

Find the acceleration at $t = 2 s$ .

Differentiate twice:

2 vx = 6t − 18t + 12,

(6)

ax = 12t − 18.

(7)

Therefore

ax(2) = 24 − 18 = 6 m ∕s2.

(8)

GRE shortcut: if position is given and acceleration is requested, go directly to the second derivative; intermediate numerical position values are irrelevant.

4 Worked GRE example 2: velocity reversal

A CAR changes velocity from $+20 m ∕s$ to $− 10 m ∕s$ in $5.0 s$ . Find the average acceleration.

−-10-−-(+20-) 2 aavg = 5 = − 6.0 m ∕s .

(9)

GRE shortcut: velocity is signed in one dimension. A reversal makes the velocity change larger than the change in speed alone.

5 GRE-speed questions

M01-03G-Q01

A ball is thrown vertically upward. At its highest point, neglecting air resistance,

(A) $v = 0$ and $a = 0$ (B) $v = 0$ and $a = − g$ (C) $v = g$ and $a = 0$ (D) $v = − g$ and $a = − g$

M01-03G-Q02

A particle moves left with $vx = − 5 m ∕s$ and $ax = − 2m ∕s2$ . The particle is

(A) speeding up (B) slowing down (C) at rest (D) moving with constant speed

M01-03G-Q03

A car moves at constant speed $10 m ∕s$ around a circle of radius $20 m$ . Its acceleration magnitude is

(A) $0$ (B) $2 0.5m ∕s$ (C) $2 5m ∕s$ (D) $2 20m ∕s$

M01-03G-Q04

On a velocity-time graph, the tangent at one instant is horizontal. The instantaneous acceleration is

(A) positive (B) negative (C) zero (D) impossible to determine

M01-03G-Q05

Frames $S$ and $′ S$ move with constant relative velocity. A particle has acceleration $2 3ex m∕s$ in $S$ . Its acceleration in $S ′$ is

(A) $0$ (B) $3ex m ∕s2$ (C) dependent on frame speed (D) opposite in direction

6 Answers and brief rationales

Q01: B. Velocity is instantaneously zero at the top, but gravitational acceleration remains downward.
Q02: A. Velocity and acceleration have the same sign, so the speed increases.
Q03: C. $ac = v2∕R = 100 ∕20 = 5 m∕s2$ .
Q04: C. Acceleration is the slope of the velocity-time graph.
Q05: B. Galilean inertial frames related by constant velocity measure the same acceleration.