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

Answer

B

Explanation

Here, Mass of the car=m
Initial velocity, u=0 (As the car starts from rest)
Final velocity `bar(v)=vbar(i)` along east
t=2s
Using, `v=u+atRightarrow v hati=0+bar(a)xx2 or bar(a)=(v)/(2)hati`
Force exerted on the car is `bar(F)=mbar(a)=(mv)/(2) hat(i)=(mv)/(2)"eastward"`
This is due to the friction on the tyres exerted by the road.