Question

Dave is riding on a straight path. After 5 s, his speed is 3.0 m/s. After 7 s, his speed is 5.0 m/s. After 10 s, his speed was 7.0 m/s. What was his acceleration from 5 s to 10 s?

Answer 1

Acceleration= change
-----------
Time
The change in velocity is 4 m/s b/c 7-3=4.

The time taken is 5 s b/c 10-5 = 5.

So its 4/5 which is
0.8 m/s^2.

You're welcome.

Answer 2

Answer : His acceleration from 5 s to 10 s will be,
`0.8m/s^2`

Solution : Given,

Initial velocity = 3 m/s

Final velocity = 7 m/s

Initial time = 5 s

Final time = 10 s

Acceleration : Acceleration is defined as the rate of change of velocity with respect to the time taken.

Formula used :

`a=(\Delta V)/(\Delta t)\\\\a=(V_(final)-V_(initial))/(t_(final)-t_(initial))`

where,

a = acceleration

`\Delta V` = change in velocity

`\Delta t` = change in time

Now we have to calculate the acceleration from 5 s to 10 s :

Put all the given values in the above formula, we get

`a=(V_(final)-V_(initial))/(t_(final)-t_(initial))`

`a=(7m/s-3m/s)/(10s-5s)=0.8m/s^2`

Therefore, his acceleration from 5 s to 10 s will be,
`0.8m/s^2`