Question

12. A train is moving with a velocity of 90 km/h. When the driver applies brakes the train
comes to halt in 10 seconds
(a) Calculate the retardation of the train.
(6) Distance covered by the train in 10 seconds.
[Ans. 2.5 m/s2, 125 ml​

Answer 1

Answer:

U(initial Velocity)=90km/h

=(90×1000)/60

=90000/60

9000/6

1500m/s

V(Final Velocity)=0m/s

T(time)=10s

A(retardation)=(V-U)/T

ii) Using the SECOND EQUTION OF MOTION S=UT+1/2AT^2

Explanation:

i) A(retardation)=(V-U)/T

a=(0-1500)/10

=-1500/10

-150m/s^2

ii)  Using the SECOND EQUTION OF MOTION S=UT+1/2AT^2

fill the values

S=(1500)*(10)+1/2*(-150)*(10)*(10)

S=15000-15000/2

S=(30000-15000)/2

S=15000/2

S=7500m

 

Answer 2

Answer:

this is your answer ✌