Question

a car is moving at a speed of 36 km per hour is brought to rest while covering a distance of 100 metres the mass of the car is 400 kg find the retarding force of the car and the time by the car tostop​

Answer 1

Answer:

first lets change unit of speed km/h into m/s

so 36 km/h=10m/s

first applying newton 1st law of motion

v=u+at

v=0

u=10m/s

so

o=10+at

so at=-10

now applying 2nd law of motion

s=ut +1/2at^2

here s=100

u=10m/s

so eqn will be

100=10t+1/2at^2

putting value of at

100=10t+1/2×(-10)t

100=10t-5t

so t=100/5

=20sec

so a=-10/20

=-.5m/s^2 here negative sign shows retardation.

now force =ma

=400×1/2 N

=200 N.

Answer 2

The retarding force of the car is 200N and the time by the car to stop is 20s.

Given :

Initial velocity of the car = 36km/h

Final velocity of the car = 0 km/h

Distance covered = 100 m

Mass of the car = 400 kg

To find :

Retarding force to stop the car

Solution :

Speed of the car in m/s = 36 × 5/18

= (2 × 5) m/s

= 10 m/s

We know that,

v^2 = u^2 + 2aS where v is the final velocity , u is the initial velocity , a is the acceleration and S is the distance covered

=> 0 = 10^2 + 2a × 100

=> 0 = 100 + 200a

=> 200a = -100

=> a = -100/200

=> a = -1/2

We also know,

v = u + at ; where v is final velocity , u is initial velocity , a is acceleration and t is time.

=> 0 = 10 + (-1/2) t

=> -10 = -t/2

=> -10 × 2 = (-t)

=> -20 = -t

=> t = 20 s

We also know,

F = ma , where F is the retarding force , m is the mass and a is the acceleration ( here : retardation )

=> F = 400 × 1/2

=> F = 200

Hence, the retarding force of the car is 200N and the time by the car to stop is 20s.

#SPJ3