Constant Acceleration - A Special Case — Question 6

Question 6

An average acceleration cannot replace two different stages

A cart begins at $x=0$ with east-positive velocity $+2\ \mathrm{m/s}$. It has constant acceleration $+2\ \mathrm{m/s^2}$ for 3 s, immediately followed by constant acceleration $-1\ \mathrm{m/s^2}$ for 4 s. Velocity is continuous at the switch; no instantaneous velocity jump is imposed.

1. Find position and velocity at the switch, then final position and velocity. Use each stage’s initial velocity and elapsed time.

2. Find total distance, average velocity and average speed over all 7 s, checking whether a reversal occurs.

3. Find the full-run average acceleration. If a different cart starts with the same initial state but keeps this average acceleration constant for all 7 s, find its final velocity and displacement.

4. Explain why the replacement matches one endpoint quantity but misses another. Why is applying $(v_i+v_f)T/2$ to the original full run unjustified?

Question 6 - Solution

An average acceleration cannot replace two different stages

(a) The first stage’s final state initializes the second.

After 3 s, $$v_1=2+2(3)=\boxed{8\ \mathrm{m/s}},\quad x_1=2(3)+\tfrac12(2)(3)^2=\boxed{15\ \mathrm{m}}.$$ The second stage begins at that position with velocity 8 m/s. Over its 4 s, $$v_f=8-4=\boxed{4\ \mathrm{m/s}},\quad \Delta x_2=8(4)-\tfrac12(4)^2=\boxed{24\ \mathrm{m}}.$$ Thus $\boxed{x_f=39\ \mathrm{m}}$.

(b) Verify the direction before equating distance and displacement.

Velocity increases from 2 to 8 m/s, then decreases only to 4 m/s. It stays positive, so no reversal occurs. Hence $$\boxed{D=39\ \mathrm{m}},\qquad \boxed{\bar v=+39/7\ \mathrm{m/s}},\qquad\boxed{\bar s=39/7\ \mathrm{m/s}}.$$

(c) The average preserves net velocity change.

The full-run average acceleration is $(4-2)/7=\boxed{2/7\ \mathrm{m/s^2}}$. A replacement with this constant acceleration would have $$v_f^{*}=2+(2/7)(7)=4\ \mathrm{m/s},$$ $$\Delta x^{*}=2(7)+\tfrac12(2/7)(7)^2=\boxed{21\ \mathrm{m}}.$$ It reaches the same final velocity but a position 18 m behind the original cart.

(d) Displacement depends on how velocities are distributed in time.

$$\Delta x=\frac{2+8}{2}(3)+\frac{8+4}{2}(4)=39\ \mathrm{m}\ne\frac{2+4}{2}(7)=21\ \mathrm{m}.$$

The original cart spends much of its run above both its initial and final velocities. The replacement increases gently from 2 to 4 m/s. Equal net velocity change does not make these histories have equal displacement.

The endpoint-velocity arithmetic mean requires a constant-acceleration interval as a sufficient justification. The original 7 s run has two distinct accelerations. Applying the formula to that whole run would silently replace its actual history by the one yielding 21 m. Apply it separately on each valid stage instead.