Another Look at Constant Acceleration — Question 2

Question 2

A definite integral gives displacement, not automatically distance

A point has $x(0)=5\ \mathrm{m}$ and constant acceleration $+2\ \mathrm{m/s^2}$ with $v(0)=-6\ \mathrm{m/s}$. Hence its velocity is $v(t)=-6+2t$ in m/s when $t$ is numerical seconds, over $0\leq t\leq4$.

The fundamental theorem of calculus gives $$x(T)-x(0)=\int_0^T v(t)\,dt.$$ Total traveled distance is $D=\int_0^T|v(t)|\,dt$. Evaluate a polynomial integral by finding an antiderivative and subtracting its lower-bound value from its upper-bound value.

1. Compute the signed integral over 0–4 s and the final position. Why is adding the initial position necessary?

2. Locate the reversal. Evaluate the distance integral by splitting it where the sign of velocity changes.

3. Compare $\left|\int_0^4v\,dt\right|$ and $\int_0^4|v|\,dt$. Explain the different order of operations physically.

4. Find both full-interval averages. Would negative displacement imply westward velocity at the final instant?

Question 2 - Solution

A definite integral gives displacement, not automatically distance

(a) Evaluate a signed change using the integral’s bounds.

An antiderivative of the numerical velocity is $-6t+t^2$. Thus $$\Delta x=[-6t+t^2]_0^4=-24+16=\boxed{-8\ \mathrm{m}}.$$ This is the change in position, so $\boxed{x(4)=5-8=-3\ \mathrm{m}}$. The integral does not retain the absolute position constant; different starting coordinates can have the same velocity history and displacement.

(b) Find where speed changes its expression.

Velocity is zero at $\boxed{t=3\ \mathrm{s}}$, negative before and positive afterward. Therefore $$D=\int_0^3(6-2t)\,dt+\int_3^4(-6+2t)\,dt =9+1=\boxed{10\ \mathrm{m}}.$$ The point moves 9 m west, then 1 m east. Its turning position is $5-9=-4\ \mathrm{m}$, consistent with finishing at $-3\ \mathrm{m}$.

(c) Cancellation depends on when the magnitude is taken.

$$D-|\Delta x|=10-8=2\ \mathrm{m}=2\min(9,1)\ \mathrm{m}.$$

$$\boxed{\left|\int_0^4v\,dt\right|=8\ \mathrm{m}},\qquad \boxed{\int_0^4|v|\,dt=10\ \mathrm{m}}.$$ Taking magnitude after integrating gives the magnitude of the net signed change, where opposing motion cancels. Taking magnitude before integrating accumulates traveled lengths in both directions without cancellation. Equality would hold for a path with no opposing traveled lengths, but not for this reversal.

(d) Averages describe the interval, not only its final event.

$$\boxed{\bar v=-8/4=-2\ \mathrm{m/s}},\qquad \boxed{\bar s=10/4=2.5\ \mathrm{m/s}}.$$ The final velocity is $v(4)=-6+8=\boxed{+2\ \mathrm{m/s}}$, eastward despite the westward net displacement. The integral records the entire interval’s signed accumulation.