Question

a train accelerates from 36 km/h to 54 km/h in 10 sec 1. find the acceleration 2.the distance travelled by the car

Answer 1

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Explanation:

GIVEN

• Initial velocity ,u = 36 kmph
• final velocity,v = 54 kmph
• time , t = 10 sec

TO FIND

• Acceleration of the train
• distance travelled by the train

EQUATIONS USED

• v = u + at
• ${v}^{2} - {u}^{2} = 2as$

First, we should convert initial velocity and final velocity into m/s

We know that 1 kmph = 5/18 m/s

Given , initial velocity = 36 kmph

= 36 × 5/18 m/s

= 10 m/s

Given, final velocity = 54 kmph

= 54 × 5/18 m/s

= 15 m/s

Now, consider the equation of motion,

i.e. v = u + at

Substitute the values of v,u and t

15 = 10 + a(10)

15 = 10 + 10a

10a = 15 - 10

10a = 5

a = 5/10

a = 0.5 m/s^2

There fore ,the acceleration of the train is 0.5 m/s^2

Now , we have to calculate the distance travelled by the car

Consider the equation of motion,

i.e.

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

Substitute

v = 15

u = 10

a = 0.5

=>

${15}^{2} - {10}^{2} = 2(0.5)s$

$225 - 100 = 1s$

$125 = s$

or

s = 125 m

There fore, the distance travelled by the train is 125m

HOPE IT HELPS YOU...