A ca travelling at ` 9 ms ^(-)` accelerates and attains a speed of `27 ms ^(-1) ` in 5s. Calculate the average acceleration and the distance travelled in 5 s.

Explanation

We are given,
At t = 0 , velocity of car ` = 9 ms ^(-1)`
t = 5 s, velocity of car `= 27 ms ^(-1).`
We are to calculate, (i) Average acceleration (ii) Distance travelled in 5 s.
Using relation , (i0 Average acceleration `= ("Change in velocity")/("Time interval") = ( 27 -9)/(5) =(18)/(5) = 3.6 ms ^(-2)`
(ii) Distance travelled `= ut + (1)/(2) at ^(2) = ( 9 xx 5 ) + (1)/(2) xx 3. 6 xx (5) ^(2) = 90 m.`