Question

A train starting from rest attains a velocity of 72km/hr in 5 mins find its acceleration and distance(assuming that it's acclration is uniform )

Answer 1

EXPLANATION.

• GIVEN

Let Train starting from rest

Therefore, u = 0 = initial velocity

Attained velocity of 72 km/hr

72 X 5 / 18 = 20 m/s = final velocity

Time = 5 minutes

60 seconds = 1 minutes

5 X 60 = 300 seconds

TO FIND ACCELERATION AND DISTANCE

1) = From newton 1 st equation of kinematics

v = u + at

20 = 0 + a(300)

300a = 20

a = 1 / 15 m/s^2

2) = From newton 2nd equation of kinematics

$s = ut + \frac{1}{2} at {}^{2}$

$s \: = (0)300 + \frac{1}{2} \times \frac{1}{15} \times 300 \times 30$

$s \: = \frac{1}{2} \times \frac{1}{15} \times 300 \times 300$

s = 3000 m