Question

moving train is brought to rest within 40 sec by applying brakes. Find the initial velocity

of train if the retardation (uniform) due to brake is 5 ms-2

.Find the distance travelled by the train

during this time. . (Ans: u = 200 m/s, s = 4000 m)​

Answer 1

Explanation:

A moving train brought to rest in=40 s(t)

Now,the acceleration of thr train after applying brake= 5m/s-²

Therefore, initial velocity (u)=a×t

=5m/s-² × 40 s

=200m/s

Total distance travelled. s = vt+½at²

=200×40+½×5(40)²

=4000m

Answer 2

The initial velocity of the train = 200 m/s

Distance travelled by train = 4000 m

Given:

Time taken = 40 sec

Retardation of the train = 5 m/s²

To find:

The initial velocity of the train.

distance travelled by train.

Solution:

• When an object is moving with a constant velocity it continues to move with that velocity until an external force is applied to that object.
• The external force may be in the form of acceleration or deacceleration.
• in this case, as constant retardation occurs therefore the train comes to a position where its velocity becomes zero.

The train is brought to rest in 40 sec

Retardation of the train = 5 m/s²

therefore,

final velocity (v) = initial velocity (u) + at

v = u + at

where,

v = 0

a = -5 m/s² (retardation)

t = 40 sec

Therefore putting the values in the equation we get

0 = u - 5 × 40

u = 5 × 40

u = 200 m/s

Distance covered in 40 sec

s = ut + 1/2 at²

s = (200 × 40) - 1/2 × 5 × (40)²

s = 4000 m

Therefore,

The initial velocity of the train = 200 m/s

Distance travelled by train = 4000 m

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