GRE Physics Companion: Constant Acceleration Motion

This companion is designed for rapid review after completing M01-05. The emphasis is not on re-deriving the equations, but on selecting the correct relation quickly, assigning signs consistently, and recognizing common test traps.

1 Fast equation triage

For one-dimensional constant acceleration, keep the five variables

v0, v, a, t, Δx

(1)

in view. Most problems give three or four of them and ask for another.

The four standard relations are

v = v0 + at,

(2)

Δx = v t + 1at2, 0 2

(3)

v2 = v2+ 2aΔx, 0

(4)

and

v0 + v Δx = ------t. 2

(5)

A fast strategy is to select the equation that contains the unknown and avoids the variable that is not supplied.

Figure 1. GRE-speed workflow: list the actual variables, choose an equation that omits the unused one, set the sign convention, and then check units and physical sense.

2 Sign and free-fall traps

A sign belongs to a component, not to a physical word such as “speeding up” or “gravity.” Choose the positive axis first.

With upward positive near Earth’s surface,

ay = − g.

(6)

At the top of a vertical throw,

vy = 0,

(7)

but

ay = − g.

(8)

On the way downward, both v_y and a_y are negative, so the object is speeding up.

Figure 2. Three common constant acceleration test traps: zero velocity does not imply zero acceleration, negative acceleration does not always mean slowing down, and return to the launch height restores the launch-speed magnitude when drag is absent.

3 Worked GRE example 1: no-time equation

A CAR moving at 20 m∕s brakes uniformly and stops in 50 m. What is the acceleration?

Time is absent, so use

v2 = v2+ 2aΔx. 0

(9)

With v = 0,

2 0 = 20 + 2a(50 ).

(10)

Thus

a = − 4.0 m∕s2.

(11)

The negative sign is expected because the positive direction was chosen along the initial motion.

4 Worked GRE example 2: vertical launch

A ball is launched upward at 19.6 m∕s. Neglect drag and use g = 9.8 m∕s². How long after launch does it return to the launch height?

The time to the top follows from

0 = v0 − gttop,

(12)

ttop = 2.0 s.

(13)

The return time is twice this value:

t = 4.0 s. return

(14)

A fast symmetry argument reaches the same result.

5 GRE-speed questions

M01-05G-Q01

A particle has v₀ = 5 m∕s and a = 3 m∕s². Its speed after 4 s is

(A) 8 m/s (B) 12 m/s (C) 17 m/s (D) 20 m/s (E) 25 m/s.

M01-05G-Q02

An object starts from rest and accelerates uniformly. If its acceleration doubles while the elapsed time is unchanged, its displacement from the start is multiplied by

(A) 1∕2 (B) 1 (C) 2 (D) 4 (E) 8.

M01-05G-Q03

A ball thrown straight upward reaches its highest point. At that instant its velocity and acceleration are

(A) both zero; (B) zero velocity and downward acceleration; (C) upward velocity and zero acceleration; (D) downward velocity and downward acceleration; (E) zero velocity and upward acceleration.

M01-05G-Q04

A car moving at 10 m/s accelerates uniformly at 2 m/s² for 5 s. Its displacement is

(A) 25 m (B) 50 m (C) 75 m (D) 100 m (E) 125 m.

M01-05G-Q05

A particle has negative velocity and negative acceleration. Its speed is

(A) increasing; (B) decreasing; (C) zero; (D) necessarily constant; (E) impossible to determine even if both signs are known.

M01-05G-Q06

For constant acceleration, which relation contains no explicit time variable?

(A) v = v₀ + at; (B) Δx = v₀t + at²∕2; (C) v² = v₀² + 2aΔx; (D) Δx = (v₀ + v)t∕2; (E) none of these.

M01-05G-Q07

An object released from rest falls for time t with constant gravitational acceleration. If the time is doubled, the distance fallen is multiplied by

(A) 2 (B) 3 (C) 4 (D) 6 (E) 8.

M01-05G-Q08

A particle’s velocity-time graph is a straight horizontal line above the time axis. The acceleration is

(A) positive constant; (B) negative constant; (C) zero; (D) increasing; (E) decreasing.

6 Answers and concise rationales

C. v = 5 + 3(4) = 17 m/s.
C. From Δx = at²∕2 when v₀ = 0, displacement is directly proportional to a.
B. The velocity is momentarily zero, but gravity remains downward.
C. Δx = 10(5) + (2)(25) = 75 m.
A. Velocity and acceleration have the same sign, so the speed magnitude increases.
C. The equation v² = v₀² + 2aΔx eliminates time.
C. From Δy = gt²∕2, doubling t multiplies the distance by four.
C. A horizontal velocity-time graph has zero slope, so a = 0.

7 What to remember under time pressure

Choose a positive direction before inserting signs.
Use the no-time relation when time is absent.
At the top of free fall, velocity can be zero while acceleration is not.
Same signs of v and a mean speeding up; opposite signs mean slowing down.
Areas and slopes from M01-04 remain valid and can provide a faster route than algebra.

References

[1] PhysicsLibrary, M01-05: Constant Acceleration Motion.

[2] OpenStax / cnxuniphysics, University Physics Volume 1, archived 2016 BCcampus clone, CC BY 4.0.

[3] University of California, Davis, Physics 9A: Classical Mechanics, Physics LibreTexts, CC BY-SA 4.0.

"GRE Physics Companion: Constant Acceleration Motion" is owned by bloftin.
(view preamble)