Question

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration is uniform, find (i) The acceleration (ii) The distance travelled by the train for attaining this velocity.

Answer 1

Answer:

The acceleration of the train is 1/15ms^-2 and  distance travelled is 3km.

Explanation:

Given :

• Initial velocity (u) => 0
• Final velocity (v) => 72km h^-1
• Time (t) => 5min = 5*60=300s

To Find :

• The acceleration
• Distance travelled by the train for attaining velocity of 72km h^-1

Solution :

1)We know that

$\Rightarrow \sf{a}=\dfrac{(v-u)}{u}$

Put their values

$\Rightarrow \sf{}\dfrac{20ms^{-1}-0ms^{-1}}{300}$

$\sf{}\because\ Acceleration\ \dfrac {1}{15}ms^{-2}$

$\sf{} 2)Use\ 2as=v^{2}-u^{-2}\ to\ find\ speed$

$\Rightarrow \sf{}s=\dfrac{v^{2}}{2a}$

Put their values

$\Rightarrow \sf{} \dfrac{(20ms^{-1})}{2\times(1\times25)}ms^{-2}$

$\Rightarrow \sf{}300m$

Change it in km

$\sf{}\because 3km$

Answer 2

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