Question

A train starting from rest moves with a uniform acceleration of 0.2 metre per second square for 5 minutes calculate the speed acquired and distance

Answer 1

Initial velocity (v)= 0m (as it starts from rest)
Acceleration (a)= 0.2m/s^2
Time (t)=5 minute
= 5×60 = 300s
Let final velocity be v

Using acceleration formula
a=(v-u)/t
0.2 = v/300
60 =v
So, final velocity was 60m/s.

Let distance be s
Using second equation of motion
s=ut+(1/2)at^2
s= (0)(300)+(1/2)(0.2)(300)^2
=9000m
So, the distance covered by the train was 9000m or 9km.

Answer 2

initial speed = 0

time taken = 5 min = 300 s

let final speed = v

then,

use kinematics equation ,

v = u + at

= 0 + 0.2 × 300 = 60 m/s

and distance travelled = S ( let )

use formula ,

S = ut + 1/2at²

= 0 + 1/2 × 0.2 × (300)²

=0.1 × 90000

=9000 m

= 9 km

hope its help you

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