A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with unifrom acceleration the force exerted on the car is .

Options

A. `(m upsilon)/(2)` eastward and is exerted by the car engine
B. `(m upsilon)/(2)` eastward and is due to the friction on the tyres exerted by the road
C. more than `(m upsilon)/(2)` eastward exerted due to the engine and overcomes the friction of the road
D. `m (upsilon)/(2)` exerted by the engine

Answer

b

Explanation

Here, mass of car `=m`
As it starts from rest, `u = 0` final velocity along east `upsilon = upsilonhati , t =2 sec`
From `upsilon = u + at`
`upsilonhati = 0 + vec (a) xx 2, vec(a) = (upsilon)/(2) hati`
`vec(F) = m vec(a) =( m upsilon)/(2) hati` ,force of car is` (m upsilon)/(2)` east ward
This is due to friction on the tyres exerted by the road .