Question

2. The engine of a train passes an electric pole with a velocity 'u' and the last compartment of the train crosses the same pole with a velocity v. Then find the velocity with which the mid-point of the train passes the pole. Assume
acceleration to be uniform.

with full solution​

Answer 1

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Answer:

$v_{m} =\sqrt{\frac{v^{2} + u^{2}}{2}}$

Explanation:

we are given that

since it is a train is not possible that first compartment has different velocity than last compartment.

initial velocity = u

final velocity = v

acceleration = a (since it is constant lets assume its a)

length of train = s (length cannot change so lets assume its s)

so it is said that after distance s its velocity changes from u to v under the influence of acceleration a.

$v^{2} - u^{2} = 2as$

a = $\frac{v^{2} - u^{2}}{2s}$

now we are asked about mid point.

the midpoint will be s/2

again use the same formula

$v_{m}^{2} - u^{2} = 2as_{m}$

insert the value of a and s

$v_{m}^{2} - u^{2} = 2 * \frac{(v^{2} - u^{2})}{2s} *  \frac{s}{2}$

$v_{m}^{2} - u^{2} = \frac{v^{2} - u^{2}}{2} = \frac{v^{2}}{2} - \frac{u^{2}}{2}$

$v_{m}^{2} = \frac{v^{2} + u^{2}}{2}$

$v_{m} =\sqrt{\frac{v^{2} + u^{2}}{2}}$