Answer to Question #94120 in Mechanics | Relativity for Jimmy

Question

An object is moving under a constant acceleration (never changes at any point). It is initially moving in the +x direction at 9 m/s. After some time the object stops and then starts moving in the -x direction. After a total time of 5 seconds, its average velocity for the entire trip is 2 m/s in the -x direction. How much distance does the object cover in this time in meters?

Answer 1

Let $t_1$ is a time of motion in the +x direction and $t_2$ is a time of motion in the -x direction.

The acceleration of an object

$$a=\frac{0-v_i}{t_1}=-\frac{9}{t_1}$$

The distance in the +x direction

$$S_1=\frac{0-v_i^2}{2a}=\frac{9}{2}t_1$$

The distance in the -x direction

$$S_2=\frac{at_2^2}{2}=-\frac{9t_2^2}{2t_1}$$

The average velocity

$$v_{ave}=\frac{S_1+S_2}{t_1+t_2}=\frac{\frac{9}{2}t_1-\frac{9t_2^2}{2t_1}}{5}=2$$

We get an equation

$$t_1-\frac{t_2^2}{t_1}=\frac{20}{9}$$

Also we have second equation

$$t_1+t_2=5$$

These equations have solution

$$t_1=\frac{45}{14}, \quad t_2=\frac{25}{14}$$

So

$$S_1=\frac{9}{2}\times \frac{45}{14}=\frac{405}{28}\:\rm m$$ $$S_2=-\frac{9\times (25/14)^2}{2\times 45/14}=-\frac{125}{28}\:\rm m$$

Finally, the total distance

$$|S_1|+|S_2|=\frac{405}{28}+\frac{125}{28}=\frac{265}{14}\:\rm m=18.9\:m$$