Question

What is t = 12.5 s to t = 37.5 s?

Answer 1

Acceleration in the time interval
`t=12.5 s` to `t=37.5 s` is `1.92 m/s^2`
Acceleration in the time interval
`t=0 s` to `t=50 s` is `0.96 m/s^2`

Explanation:

Acceleration is not changing between 12.5 s to 37.5 s
Initial velocityin m/s is `=-24`
Final velocity in m/s is`=24`
Change in velocity is `24-(-24)`
`=24+24`
Change in velocity in m/s is`48`
Time interval in s is =`37.5-12.5`
Time interval in s is `=25`
Rate of change in velocity is
`48/25`
Acceleration in m/s^2 is `1.92`
Acceleration in the time interval
`t=12.5 s` to `t=37.5 s` is `1.92 m/s^2`
The change in velocity from 0 to 50 s remains the same.
But the time interval is increased.
Hence the acceleration is reduced

Rate of change in velocity is
`48/50`
Acceleration in m/s^2 is `0.96`
Acceleration in the time interval
`t=0 s` to `t=50 s` is `0.96 m/s^2`

Answer 2

Average acceleration = change in velocity/total time taken for the change

Now, `t=12.5` means `12.5 /2.5=5` rooms along time axis in the graph.

At that point value of velocity is `8` rooms along negative `v` axis,so value of velocity at `12.5 s` is `-8*3=-24 m/s`

Similarly, `t=37.5` means `37.5/2.5=15` rooms along time axis.

at that point value of velocity is `8` rooms along positive `v` axis,so velocity at `37.5 s` is `8*3 =24 m/s`

So,average acceleration in this time interval is `(24-(-24))/(37.5-12.5) =1.92 m/s^2`

Similraly, `t=50` means `50/12.5=20` rooms along time axis.

so,corresponding velocity value is `8` rooms along positive `v` axis.

And, at `t=0` velocity is `8` rooms along negative `v` axis,so its value will be `-24 m/s`

So,the velocity value is `24 m/s`,

So,average acceleration in this time interval is =`(24-(-24))/(50-0) =0.96 m/s^2`