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

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

Step-by-step 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=(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=(v-v_0)/(t-t_0)=((-8m/s)-(5m/s))/((10s)-(8s))=(-13m/s)/(2s)=-6.5m/s^2`

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