Constant Acceleration - A Special Case — Question 3

Question 3

One position can correspond to two different velocities

A point starts at $x=0$ with $v_i=+12\ \mathrm{m/s}$ and maintains $a=-4\ \mathrm{m/s^2}$ for 6 s. It is allowed to reverse and continue west. Thus, with $t$ numerical seconds, $$x(t)=12t-2t^2\ \mathrm{m},\qquad v(t)=12-4t\ \mathrm{m/s}.$$

1. Find every time in the specified interval when $x=16\ \mathrm{m}$. Give the velocity at each event.

2. Eliminate time using $v^2=v_i^2+2a(x-x_i)$. Why does taking only the positive square root miss one of the events?

3. Find the turning time and position. Compute distance traveled from departure to each visit to 16 m, and compare with displacement.

4. Find average speed and average velocity from departure to the second visit. Does negative acceleration imply decreasing speed throughout that interval?

Question 3 - Solution

One position can correspond to two different velocities

(a) Keep all roots inside the model’s time domain.

The position equation gives $12t-2t^2=16$, or $(t-2)(t-4)=0$. Thus $$\boxed{t=2\ \mathrm{s},\ v=+4\ \mathrm{m/s}},\qquad \boxed{t=4\ \mathrm{s},\ v=-4\ \mathrm{m/s}}.$$ The first visit is eastward; the second is westward.

(b) Squaring removes direction information.

$$v=\pm\sqrt{v_i^2+2a\Delta x}=\pm4\ \mathrm{m/s},\qquad t=\frac{v-v_i}{a}\in\{2,4\}\ \mathrm{s}.$$

At the target, $v^2=144+2(-4)(16)=16\ \mathrm{m^2/s^2}$. Therefore $v=\pm4\ \mathrm{m/s}$. The squared equation alone does not select which part of the trajectory is intended. The corresponding times and direction information resolve the two branches; both are valid here.

(c) Split the distance at the reversal.

Setting $v=0$ gives $\boxed{t=3\ \mathrm{s}}$, with $\boxed{x=18\ \mathrm{m}}$. The first visit has distance 16 m and displacement $+16\ \mathrm{m}$. Before the second visit, the point travels 18 m east and then 2 m west: $$\boxed{D_{0\to4}=20\ \mathrm{m}},\qquad\boxed{\Delta x_{0\to4}=+16\ \mathrm{m}}.$$

(d) Different numerators produce different averages.

$$\boxed{\bar s=20/4=5\ \mathrm{m/s}},\qquad\boxed{\bar v=16/4=+4\ \mathrm{m/s}}.$$ Speed decreases before 3 s while velocity is positive, then increases while both velocity and acceleration are negative. The acceleration remains negative throughout; its sign alone does not determine the speed trend.