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

The average acceleration of the particle during this 9.1 s interval is
`a=2.31 ms^(-2)`.

Step-by-step explanation:

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

`a_(vg)=(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)=(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)`.