A2.1 × 103
kg car accelerates from rest under
the action of two forces. one is a forward
force of 1141 n provided by traction between
the wheels and the road. the other is a 911 n
resistive force due to various frictional forces.
how far must the car travel for its speed to
reach 2.9 m/s?
answer in units of m.

First solve the acceleration of the car using the formula
Fnet = ma Where Fnet is the net force acting on the body M of the object A is the acceleration  
( 1141 – 911) = (2.1 x10^3) a
a = 0.1095 m/s2  

a = v/t
0.1095 = 2.9 /t
t = 26.4783 s  

d = vt
d = ( 2.9 m/s) ( 26.4783 s)
d = 76.7870 m

Ithink the answer is b