Question

A train starting from rest attains a velocity of 72 Km/h (20 m/s) in 5 minutes.

Calculate its acceleration and distance covered.​

Answer 1

Answer:

initial velocity=0 m/s

final velocity=20 m/s

Time taken =5 minutes=300 seconds

Acceleration=(20 m/s-0 m/s)/300=20 m/s /300=1/15 m/s²

2as=v²-u²

2x1/15xs=(20 m/s)²-(0 m/s)²=400 m/s²

s=400 m/s²x15/2=300 metres

Answer 2

Answer:

Acceleration = 900 km/h

Distance = 2.88 km

Explanation:

Given:

initial velocity = 0 km/h

final velocity = 72 km/h

time = 5 min = 0.08h

Acceleration = ?

acceleration = final velocity - initial velocity/time

acceleration = 72-0/0.08 =72 / 0.08 = 900

When using equations of motions, distance = displacement and velocity = speed

so,

displacement = initial velocity*time + half(acceleration*time^2)

displacement or distance = 0*0.08 + 1/2 (900*0.08^2)

displacement or distance = 0 + 5.76/2

distance = 2.88 km

I am not really sure about my distance calculation.

Thank you