Acceleration — Question 9

Question 9

Equal accelerations do not determine equal velocities

Two markers move on the same east-positive straight line with exact position histories $$x_A(t)=t+\tfrac13t^3,\qquad x_B(t)=10-3t+\tfrac13t^3,\qquad0\leq t\leq3.$$ Here $t$ is numerical seconds and positions are numerical metres. The markers may pass through each other; no interaction is imposed. Differentiate to obtain velocity and acceleration.

1. Show that their acceleration histories are identical. At 1 s, compare their velocities, speeds, and whether their speeds are increasing or decreasing.

2. Find $v_B-v_A$ and $x_B-x_A$. Are their velocities ever equal? When do they meet, and what are their velocities at that meeting?

3. Explain how two markers with identical accelerations can have different directions or different speed changes. Which information present in the position functions disappears after differentiating twice?

4. More generally, suppose two differentiable velocity histories have equal acceleration throughout a connected interval. Using the fact that a differentiable function with zero derivative is constant, explain which initial data are needed to make their velocities agree everywhere, and which further data make their positions agree everywhere.

Question 9 - Solution

Equal accelerations do not determine equal velocities

(a) Differentiate each position history twice.

$$v_A=1+t^2,\quad v_B=-3+t^2\quad(\mathrm{m/s}),\qquad \boxed{a_A=a_B=2t\ \mathrm{m/s^2}}.$$ At 1 s, $v_A=+2$ and $v_B=-2\ \mathrm{m/s}$; both speeds are 2 m/s and both accelerations are $+2\ \mathrm{m/s^2}$. A moves east and speeds up; B moves west and slows down. The same acceleration acts on different signed velocities.

(b) The velocity difference is constant, while the position difference varies.

$$\boxed{v_B-v_A=-4\ \mathrm{m/s}},\qquad \boxed{x_B-x_A=(10-4t)\ \mathrm{m}}.$$ The velocities never agree. Positions agree at $\boxed{t=2.5\ \mathrm{s}}$, when $$\boxed{v_A=7.25\ \mathrm{m/s},\qquad v_B=3.25\ \mathrm{m/s}}.$$ A shared position does not remove the persistent velocity difference.

(c) Acceleration alone omits initial position and velocity information.

The functions differ by a constant term and a term linear in time. The first derivative removes the constant position offset; the second removes the constant velocity difference. Their identical accelerations therefore do not force identical velocities, directions, or speed changes. At 1 s the two velocity signs produce opposite speed trends even though the acceleration signs agree.

(d) Supply the missing initial data in the correct order.

$$a_B-a_A=0\Rightarrow v_B-v_A=C_1\Rightarrow x_B-x_A=C_1t+C_2.$$

If $a_B-a_A=0$ everywhere, then the derivative of $v_B-v_A$ is zero, so that velocity difference is constant. Equal velocities at one time make this constant zero, giving equal velocities throughout the interval.

With equal velocities everywhere, the derivative of $x_B-x_A$ is zero, so the position difference is constant. Equal positions at one time then make that constant zero. Thus matching acceleration histories plus one shared initial velocity ensures matching velocities; adding one shared initial position ensures matching positions. Equality of acceleration alone supplies neither initial condition.