Question

At a certain time a particle had a speed of 46 m/s in the positive x direction, and 9.1 s later its speed was 67 m/s in the opposite direction. What was the average acceleration of the particle during this 9.1 s interval

Answer 1

Answer:

The average acceleration of the particle during this 9.1 s interval is  $a=2.31 ms^{-2}$.

Explanation:

The expression for the average acceleration of the particle is as follows;

$a_{vg}=\frac{v-u}{t'-t}$

Here, $a_{vg}$ is the average acceleration, v is the final speed, u is the initial speed, t is the initial time and t' is the final time.

It is given in the problem that at a certain time a particle had a speed of 46 m/s in the positive x-direction, and 9.1 s later its speed was 67 m/s in the opposite direction.

Put $u=46 ms^{-1}$, $v=67 ms^{-1}$, t= 0 and t'=9.1 s in the above expression.

$a_{vg}=\frac{67-46}{9.1-0}$

$a=2.31 ms^{-2}$

Therefore, the average acceleration of the particle during this 9.1 s interval is  $a=2.31 ms^{-2}$.