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**Hint:**It is given that the train undergoes retardation when the brakes are applied. Make use of the kinematic equation to find the velocity of the train after 10 seconds. Then use the same equation to find the time taken to come to rest.

**Formula used:**

$v=u+at$

**Complete answer:**

It is given that the train is initially moving with a velocity of 90km/h. Then the brakes are applied due to which the train undergoes retardation. When a body is undergoing retardation, its velocity decreases with time. In other words, the body is moving with a negative acceleration.

It is given that the train is retarding at the rate of 0.5$m{{s}^{-2}}$. Therefore, $a=-0.5m{{s}^{-2}}$.

(i) Let us find the velocity of the train after 10 seconds. For this, we will use the kinematic equation $v=u+at$ ….. equ-(i),

where v is the velocity the moving body at time t, u is its velocity at time t=0 and a is its acceleration.

In this case, $u=90km{{h}^{-1}}=90\left( {{10}^{3}}m \right){{\left( 3600s \right)}^{-1}}=25m{{s}^{-1}}$

$t=10s$ and $a=-0.5m{{s}^{-2}}$.

Substitute the values in equ-(i).

$v=25+(-0.5)(10)=25-5=20m{{s}^{-1}}$.

This means that the velocity of the train after 10s of applying the brakes is $20m{{s}^{-1}}$.

(ii) Let the time taken for the train to come to rest be t’. Now, v=0. After substituting the values in (i) we get that

$0=25+(-0.5)(t')$

$\Rightarrow 0.5t'=25$

$\Rightarrow t'=50s$

This means that the train comes to rest after 50 seconds.

**Note:**

Do not confuse between acceleration and retardation.

Actually, retardation is an acceleration. When we say that a body is retarding, it means that its velocity is decreasing with time and therefore, it is accelerated in the opposite direction of its velocity.

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