Question

A vehicle of 1000 kg is moving at a speed of 72 km/hr. It is stopped by applying brakes within 5 seconds. Calculate the retardation of the vehicle, the distance travelled by the vehicle after applying the brakes in 5 seconds and the amount of retarding force to stop the vehicle.
[Ans: -4 m/s2,50 m, -4000N]​

Answer 1

Answer:

-4m/s,-4000N

Explanation:

lets take retardation as nagetive acceleration

as you know acceleration is equals to v minus u divided by t

so here you is equals to to 72 kilometre per hour and b is equal to zero v is equal to zero because it applies break and time is equals to 5 seconds so after solving we may get -4 metre per second squareso now we have to calculate the force force is equals to mass in to acceleration mass is given 1000 kg and acceleration we took out - 4 metre per second square so after solving we get - 4000 Newton

Answer 2

Initial velocity(u) = 72*1000/60*60

= 20m/s

mass(m)= 1000kg

time(t)=5second

final velocity(v)=0m/s

retradation(-a)=v-u/t

( -a)=-20/5

(-a)= -4m/s^2

distance(s)=ut+1/2at^2

(s)= 20*5+1/2*(-4)*5*5

(s)=50m

force(f)=ma

(f)=1000*(-4)

(f)=-4000N