Question

A train was moving at the rate of 15 m s-1 when brakes were applied. It came to rest within a distance of 225 m. Calculate the retardation produced in the train.​

Answer 1

Explanation:

u=15

v=0

s=225m=0.225km

from equation,

v^2-u^2=2as

0^2-(15)^2 =2×a×0.225

-225\0.225=2a

a=-500

Answer 2

The retardation produced in the train $0.5 m/sec^{2}$.

Given,

Speed at which the train is moving=15 m/sec

Distance that the train covers before coming to rest=225 m.

To find,

the retardation produced in the train.​

Solution:

• When the train comes at rest, it final velocity becomes zero, v=0.
• Retardation=-(acceleration).
• The equation of motion that is used to be used here is:
• $v^{2} =u^{2} +2as$.
• where, v-final speed, u-initial speed, a-acceleration and s-distance.

The acceleration of the train will be:

$v^{2} =u^{2} +2as\\0^{2} =15^{2} +2(a)*225\\0=225+450a\\-225=450a\\a=\frac{-225}{450} \\a=-0.5m/sec^{2} .$

The retardation produced in the train will be:

$Retardation=-(acceleration)\\Retardation=-(-0.5)m/sec^{2} \\Retardation=0.5 m/sec^{2} .$

Hence, the retardation produced by the train is $0.5 m/sec^{2}$.

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