Question

a train starting from rest attains a velocity of 100km/hr in 10 minutes. If acc. is uniform find the acc. and distance covered by Train?​

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Final velocity ,v = 100km/h = 100×5/18 = 27.77m/s
• Time taken ,t = 10min = 10×60 = 600s

To Find:-

• Acceleration ,a

Solution:-

We have to calculate the acceleration of the train. As we know that acceleration is defined as the rate of change in velocity.

• a = v-u/t

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

$:\implies$ a = 27.77-0/600

$:\implies$ a = 27.77/600

$:\implies$ a = 0.046m/s²

• Hence, the acceleration of the train is 0.046m/s².

Now, calculating the distance covered by the train. Using 3rd equation of motion

• v² = u² + 2as

substitute the value we get

$:\implies$ 27.77² = 0² + 2×0.046 × s

$:\implies$ 771.1729 = 0 + 0.092 ×s

$:\implies$ 771.1729 = 0.092 ×s

$:\implies$ s = 771.1720/0.092

$:\implies$ s = ≈ 8382 m

• Hence, the distance covered by the train is 8382 metres (approx).

Answer 2

Given :-

A  train starting from rest attains a velocity of 100km/hr in 10 minutes

To Find :-

Acceleration

Distance covered

Solution :-

We know that

v = u + at

1 km/h = 5/18 m/s

100 × 5/18 = 500/18 = 27.77 m/s

10 min = 600 sec

27.77 = 0 + a(600)

27.77 = 0 + 600a

27.77 = 600a

27.77/600 = a

0.045 = a

Now

v² - u² = 2as

(27.77)² - (0)² = 2(0.045)(s)

771.17 - 0 = 0.9s

771.17 = 0.09s

771.17/0.09 = s

8568 = s