Question

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.

Answer 1

Explanation:

mass of car, m = 2.7 x 10^3 kg

Forward force, F = 1157 N

friction force, f = 902 N

initial velocity, u = 0 m/s

final velocity, v = 3.6 m/s

According to Newton's second law

F - f = m x a

where a is the acceleration of the car

1157 - 902 = 2.7 x 10^3 x a

a = 0.094 m/s²

Use third equation of motion

v² = u² + 2as

3.6 x 3.6 = 0 + 2 x 0.094 x s

s = 68.94 m

Answer 2

The car covers a distance of 68.93 meters.

Explanation:

Given that,

Mass of the car,
`m=2.7* 10^3\ kg`

Forward force acting on the car,
`F_1=1157\ N`

Force of friction acting on the car,
`f=902\ N`

The net force acting on the car is :

`F=F_1-f\\\\F=1157-902\\\\F=255\ N`

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,

`F=ma\\\\a=(F)/(m)\\\\a=(255)/(2.7* 10^3)\\\\a=0.094\ m/s^2`

Now using third equation of motion as :

`v^2-u^2=2ad\\\\d=(v^2)/(2a)\\\\d=((3.6)^2)/(2* 0.094)\\\\d=68.93\ m`

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