1.

Acceleration of a car in motion is 1.5 m s^(-2). How much is the increase in velocity in 4s?

Answer»

`6 ms^(-1)`
`4 ms ^(-1)`
`4 ms ^(-1)`
`2.66 ms ^(-1)`

Solution :ACCELERATION`= ("CHANGE in velocity")/("Time interval")`
`therefore` Change in velocity (increase)
= Acceleration `xx` Time interval
`= (1.5 ms ^(-2)) xx (4S)`
`=6 ms ^(-1)`


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