An engine delivered power of 30 kW to a car of mass 1250 kg and velocity of 30 m/s . If the friction force of surface is 750N what will be the maximum acceleration of car ?

Options

A. `1/3ms^(-2)`
B. `1/4 ms^(-2)`
C. `1/5 ms^(-2)`
D. `1/6 ms^(-2)`

Answer

C

Explanation

P = 30 kW , m = 1250 kg , v = 30 m/s
Resistive force = 750 N
Force exerted by engine ,
`F =P/v =(30,0000)/30 = 1000 N `
maximum acceleration produced
` = ("Force exerted by engine " - " Resistious force ")/("mass of car")`
` a = (1000- 750)/1250 = 250/1250 =1/5 m//s^(2)`