A car weighing 19600N is moving with a speed of 30 m/sec on a level road. If it is brought to rest in a distance of 100 m. Find the average distance friction force acting on it

F = -8820 N

Explanation:

Given that,

The weight of a car, W = 19600 N

Initial speed of the car, u = 30 m/s

It is brought to rest, final velocity, v = 0

Distance, d = 100 m

We need to find the average friction force acting on it.

Firstly we find the acceleration of the car using third equation of motion. Let it is a.

Average frictional force,

F = ma

m is mass,

F = 1960 kg × -4.5 m/s²

= -8820 N

So, the average friction force acting on it is 8820 N.

we need to know the time interval during which the force acts. impulse of force = momentum change

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s