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A car accelerates uniformly from 18 km h^(-1 )to 36 km h^(-1) in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time. |
Answer» Solution :Here, `u = 18 km h ^(-1) = 5 ms ^(-1)` ` v = 36 km h ^(-1) = 10 ms ^(-1)` `(i)a = (v - u)/(t)` `=(10 ms ^(-1) - 5 ms ^(-1))/( 5S)= 1 ms ^(-2)` (II) `s = ut + 1/2 at ^(2)` `= ( 5m s ^(-1)) xx (5s ) + 1/2 xx (1m s ^(-2)) xx (5s) ^(2)` `= 25 m + 12.5 m` `= 37. 5m` The acceleration of the car is `1 ms ^(-2)` and the DISTANCE covered is `37.5 m.` |
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