A 2.7 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a 902 N resistive force due to various frictional forces. How far must the car travel for its speed to reach 3.6 m/s? Answer in units of m.

The car covers a distance of 68.93 meters.

Step-by-step explanation:

Given that,

Mass of the car,

Forward force acting on the car,

Force of friction acting on the car,

The net force acting on the car is :

Initial speed of the car, u = 0

Let d is the distance covered by the car so that its speed reaches to 3.6 m/s.

Force,

Now using third equation of motion as :

So, the car covers a distance of 68.93 meters. Hence, this is the required solution.

a negative exponent is translated as follows:

so, we have

24+x=y

step-by-step explanation: