Question

A train starts from rest and accelerates uniformly at
a rate of 2 m s-2 for 10 s. It then maintains a
constant speed for 200 s. The brakes are then
applied and the train is uniformly retarded and
comes to rest in 50 s. Find : (i) the maximum
velocity reached, (ii) the retardation in the last
50 s, (iii) the total distance travelled, and (iv) the
average velocity of the train.​

Answer 1

Answer:

Explanation:

Total distance covered(s) = Distance during acceleration(s  

1

​  

) +  distance during uniform motion(s  

2

​  

) + distance during retardation(s  

3

​  

)

For acceleration,

v=u+a∗t

v  

max

​  

=2×10=20 m/s

s=ut+(1/2)at  

2

 

s  

1

​  

=(1/2)×2×10  

2

 

    =100 m

For uniform motion,

s=vt

s  

2

​  

=20×200

    =4000 m

During retardation,

v=u+at

0=20+50t

  =−0.4 m/s  

2

 

s  

3

​  

=ut+(1/2)at  

2

 

    =20×50+(1/2)×(−0.4)×50  

2

 

    =500 m

Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m

Total time taken, t=t  

1

​  

+t  

2

​  

+t  

3

​  

=260 s

Average velocity, v  

avg

​  

=  

Total Time

Total Distance

​  

 

                                     =4600/260=17.69 m/s