A train is running at full speed when brakes are applied. In the first minute it travels 8 km and in the next minute it travels 3 km. initial speed of the train is

Open in App

Solution

displacement at nth minute is given by

Snt = u+ a ( n- 1/2) ,

where 'u' is the initial velocity , 'a' is the acceleration

8= u + a(1-1/2)....(1)

3= u +a(2-1/2)....(2)

solvingv(1) and (2)

u=21/2 km/min
= 630 km/hr
= 175 m/s

Suggest Corrections

Similar questions

Q. A train is running at full speed when brakes are applied. In the first minute it travels

8 k m

, and in the next minute it travels

3 k m

. Assuming constant retardation, initial speed of the train is

Q. Train A crosses train B, which is stationary, in 90 seconds. After a few minutes, train B travels at a uniform speed and crosses train C, running at twice the speed in the opposite direction, in 30 seconds. If the trains are of equal length, what is the ratio of their speeds?

Q. A train starts from rest and moves with a constant acceleration of

2.0 m s^{- 2}

for a half minute. The brakes are then applied and the train domes to rest in one minute. Find the position(s) of the train at half the maximum speed.

Q. A train travels

360 k m

with uniform speed. The speed of the train is increased by

5 k m / h r

, it takes

48

minutes less to cover the same distance. Find the initial speed of the train.

A train starts from rest and moves with a constant acceleration of 2.0 $\frac{m}{s^{2}}$ fbr half a minute. The brakes are then applied and the train comes to rest in one minute. Find.

(a) the total distance moved. by the train,

(b) the maximum speed attained by the train and

Join BYJU'S Learning Program

A train is running at full speed when brakes are applied. In the first minute it travels 8 km and in the next minute it travels 3 km. initial speed of the train is

displacement at nth minute is given by Snt = u+ a ( n- 1/2) , where 'u' is the initial velocity , 'a' is the acceleration 8= u + a(1-1/2)....(1) 3= u +a(2-1/2)....(2) solvingv(1) and (2) u=21/2 km/min = 630 km/hr = 175 m/s

displacement at nth minute is given by

Snt = u+ a ( n- 1/2) ,

where 'u' is the initial velocity , 'a' is the acceleration

8= u + a(1-1/2)....(1)

3= u +a(2-1/2)....(2)

solvingv(1) and (2)

u=21/2 km/min
= 630 km/hr
= 175 m/s