Question

a train accelerate from 36km/hr to 72km/hr on covring a distance of 50cm . calculate the acceleration of the train and the time taken to cover the distance 50cm​

Answer 1

correct question :

A train accelerate from 36km/hr to 72km/hr on covring a distance of 50m . calculate the acceleration of the train and the time taken to cover the distance 50m

Answer

Given :

Initial velocity of the train = 36 km/hr

Final velocity of the train = 72 km/hr

Distance covered = 50 m

To find :

the acceleration of the train and the time taken to cover the distance of 50m.

Solution :

Firstly we have to convert km/hr to m/s

1 km/hr = 5/18 m/s

72km/hr = 72 × 5/18 = 20m/s

36km/hr = 36 × 5/18 = 10m/s

As we are provided with initial velocity, final velocity and distance covered we can use third equation of motion .i.e.,

v²- u² = 2as

here,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance covered

Now, substituting all the given values in the equation,

➠ v² - u² = 2as

➠ (20)² - (10)² = (2)(a)(50)

➠ 400 - 100 = 100a

➠ 300 = 100a

➠ a = 300/100

➠ a = 3m/s²

thus, the acceleration of the train is 3m/s²

Now, using first equation of motion .i.e.,

➠ v = u + at

➠ 20 = 10 + (3)(t)

➠ 20 - 10 = 3t

➠ 10 = 3t

➠ t = 10/3

➠ t = 3.3 sec$.$

thus, the time taken by the train is 3.3sec