Question

At t=10 ~\text{s}t=10 s, a particle is moving from left to right with a speed of 5.0 ~\text{m/s}5.0 m/s. At t=20 ~\text{s}t=20 s, the particle is moving right to left with a speed of 8.0 ~\text{m/s}8.0 m/s. Assuming the particle’s acceleration is constant, determine its average acceleration. Assume positive is on the right.

Answer 1

Answer:

$a=6.5m/s^2$ to the left.

Explanation:

We can use the equation $v=v_0+a(t-t_0)$ where v is the velocity at time t and $v_0$ the velocity at $t_0$. Since we want the acceleration we write this equation as:

$a=\frac{v-v_0}{t-t_0}$

Considering the direction to the right as the positive one, we have $v_0=+5m/s$ at $t_0=10s$, and $v=-8m/s$ at $t_0=8s$, so we substitute:

$a=\frac{v-v_0}{t-t_0}=\frac{(-8m/s)-(5m/s)}{(10s)-(8s)}=\frac{-13m/s}{2s}=-6.5m/s^2$

Where the minus sign indicates it is directed to the left.