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Question

A train starts from rest and moves with a constant acceleration of $$2.0\ m/s^{2}$$ for half a minute. the brakes are then applied and the train comes to rest in one minute. Find
the total distance moved by the train,


Solution

Train starts from rest, hence the initial velocity u = 0.

It moves with acceleration = 2m/s2 for half minute (30 seconds).

Distance covered in this time interval is given by:

$$S = ut + ½ at^2$$

$$= 0 + ½ \times 2 \times 30 \times 30$$

$$= 900m$$

Velocity attained by this acceleration after 30 seconds:

$$v = u + at$$

$$=> v = 0 + 2 x 30$$

$$=> v = 60 m/s$$

From this velocity, brakes are applied and train comes to rest in 60 seconds.

The retardation is given by:

$$v = u – at$$

$$=> 0 = 60 – a \times 60$$

$$=> a = 1 m/s^2$$

Distance covered in this time:

$$V2= u2 + 2aS$$

$$=> 0 = (60)2 + 2 (-1) S$$

$$=> 0 = 3600 – 2S$$

$$=> S = 3600/2 = 1800m$$.

So, total distance moved =$$900m + 1800m = 2700m$$.

Maximum speed of the train$$ = 60 m/s$$.

Position of the train at half its maximum speed.

Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s.

(I) When the train is accelerating with an acceleration of 2 m/s,

time at which speed = 30m/s is:

$$v = u + at$$

$$=> 30 = 0 + 2 x t$$

$$=> t = 15s$$

At 15s, distance covered from origin is:

$$S = ut + ½ at^2$$

$$= 0 + ½ \times 2 \times 15 \times 15$$

$$= 225m$$

(II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 ,  time at which the speed reaches 30 m/s is:

$$v = u – at$$

$$=> 30 = 60 – 1xt$$

$$=> t = 30s$$

At 30s, distance covered is:

$$S = ut – ½ at^2$$

$$= 60 x 30 – ½ x 1 x (30)2$$

$$= 1800 – (15 x 30)$$

$$= 1800 – 450$$

$$= 1350m$$ (from the initial 900m covered).

So, distance from origin $$= 900 + 1350m = 2250m$$.


Physics

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