Question

15.
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 :-

1) Maximum velocity = 20 m/s

2) Total distance travelled = 4.6 km

3) Average velocity = 17.70 m/s

Solution:-

Case :-1

when train accelerated.

u = 0 m/s

a = 2 m/s²

t = 10 s

1)The maximum velocity reached is given by :-

→$v = u + at$

→$v = 0 + 2 \times 10$

→$v = 2 \times 10$

→$v = 20 m/s$

Hence,

Maximum velocity will be 20 m/s.

Distance travelled by train :-

→$2as = v^2 - u^2$

→$2 \times 2 \times s = (20)^2 -(0)^2$

→$4s = 400$

→$s = \dfrac{400}{4}$

→$s = 100 m$

Distance covered will be 100 m.

Case :- 2

v = 20 m/s

Since, velocity is constant.

So, acceleration will be 0.

a = 0 m/s²

t = 200 s

Distance travelled by train :-

→$s' = v \times t$

→$s' = 20 \times 200$

→$s' = 4000 m$

Distance travelled will be 4000 m.

Case :- 3

Now, train started retardation

t = 50 s

u = 20 m/s

v = 0 m/s

Retardation produced in train will be :-

→$a = \dfrac{v-u}{t}$

→$a = \dfrac{0 - 20}{50}$

→$a = \dfrac{-20}{50}$

→$a = -0.4 m/s^2$

hence, Retardation produced in train in last 50s will be -0.4 m/s².

Distance travelled before come to the rest.

→$2as = v^2 - u^2$

→$2 \times -0.4 \times s" = (0)^2 -(20)^2$

→$-0.8 s" = - 400$

→$s" = \dfrac{-400}{-0.8}$

→$s" = 500 m$

3) Total distance travelled by train :-

→$S = s + s' + s"$

→$S = 100 + 4000 + 500$

→$S = 4600 m$

→$S = 4.6 km$

Total distance travelled is = 4600 m

Total time taken = 10 + 200 + 50

= 260s

Average velocity is given by :-

→$A_v = \dfrac{\text{Total distance travelled}}{\text{Total time taken}}$

→$A_v = \dfrac{4600}{260}$

→$A_v = 17 .70 m/s$

hence, Average velocity will be 17.70 m/s.