GRE Physics Companion: Variable Acceleration Motion

This companion is designed for rapid review after M01-06. The emphasis is on recognizing whether a derivative, an area, or the chain rule relation a = v dv∕dx is the fastest route.

1 Fast triage

If acceleration is given as a function of time,

∫ Δv = a(t) dt.

(1)

If velocity is then needed for displacement,

∫ Δx = v(t)dt.

(2)

If acceleration is supplied as a function of position and time is absent, try

dv a = v --. dx

(3)

If a graph is supplied, remember that area under a(t) gives change in velocity, while area under v(t) gives displacement.

Figure 1. GRE speed triage for variable acceleration: identify which variable the acceleration depends on before choosing the calculus relation.

2 Common traps

A variable acceleration cannot be replaced by one instantaneous value in a constant acceleration formula unless the problem explicitly justifies that approximation.

The area under an acceleration time graph is Δv, not v itself. Initial velocity must still be added.

When an integrated equation gives v², the square root gives speed magnitude. The physical direction determines the sign of velocity.

Figure 2. Two recurring test traps: integration requires the initial value, and a result for v² does not by itself determine the sign of velocity.

3 Worked GRE example 1: area under acceleration

A particle has v(0) = 4 m/s. Its acceleration rises linearly from 0 to 6 m/s² during the first 3 s. Find v(3).

The change in velocity is the triangular area under the acceleration time graph:

Δv = 1-(3)(6 ) = 9 mm ∕s. 2

(4)

Therefore

v(3) = 4 + 9 = 13 mm ∕s.

(5)

4 Worked GRE example 2: position dependent acceleration

A particle has

a(x) = − 2x.

(6)

It starts from rest at x₀ = 2 m. What is its speed when it reaches x = 0?

Use

v dv = − 2x dx.

(7)

Integrating,

∫ 1 2 0 2v = − 2xdx = 4. 2

(8)

Thus

2 v = 8,

(9)

so the speed is

√ -- |v| = 2 2 mm ∕s.

(10)

5 GRE speed questions

M01-06G-Q01

A particle has a(t) = 4t m/s² and v(0) = 1 m/s. Its velocity at t = 2 s is

(A) 5 m/s (B) 8 m/s (C) 9 m/s (D) 12 m/s (E) 17 m/s.

M01-06G-Q02

The signed area under an acceleration time graph between t₁ and t₂ equals

(A) displacement; (B) distance; (C) change in velocity; (D) average velocity; (E) jerk.

M01-06G-Q03

For one-dimensional motion with acceleration specified as a(x), which identity is most useful for eliminating time?

(A) a = dx∕dt; (B) a = v dv∕dx; (C) v = ada∕dx; (D) v = xdx∕dt; (E) a = d²v∕dx².

M01-06G-Q04

A particle satisfies a(t) = 6t m/s² and starts from rest. Which expression gives its velocity?

(A) v = 6t; (B) v = 3t²; (C) v = 6t²; (D) v = 2t³; (E) v = t³.

6 Answers and rationales

Q01: (C). Integrate: Δv = ∫ ₀²4tdt = 8 m/s, then add v₀ = 1 m/s.

Q02: (C). Since a = dv∕dt, integration over time gives Δv.

Q03: (B). The chain rule gives a = (dv∕dx)(dx∕dt) = v dv∕dx.

Q04: (B). v = ∫ 6tdt = 3t² + C and v(0) = 0 gives C = 0.

References

[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005. Used as a scope and notation reference.

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

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