A constant power of 7 W is supplied on a toy car of mass 15 kg. The distance travelled by the car when its velocity increases from `3ms^(-1)` to `5 ms^(-1)` is

Options

A. 56m
B. 7 m
C. 61 m
D. 70 m

Answer

D

Explanation

Given, `P = 7 W` , mass m = 15 kg,
`v_(i) = 3ms^(-1)` and `v_(f) = 5 ms^(-1)`
From work-energy theorem,
Work done `= (1)/(2)m(v_(f)^(2) - v_(i)^(2))`
`rArr = (1)/(2) xx 15(25 - 9)`
`rArr W = 120 J`
So, the time of work done,
`t = ("work")/("power") = (120)/(7) = 12.14 s`
Hence, the acceleration,
`a = (v_(f) - v_(i))/(t) = (5-3)/(17.14) = 0.116 m//s^(2)`
Distance travelled by car,
`D = (v_(f)^(2) - v_(i)^(2))/(2a) = (5^(2) - 3^(2))/(2 xx 0.116) = 68.96 ~~ 70 m`
Hence, the correct option is (d).