Question

A train starting from rest attains a velocity of 72 km/h in 5 minutes assuming the acceleration is uniform, find (1) The acceleration (2) The distance travelled by the train for attaining this velovity

Answer 1

hey dear
here is your answer

given data :

initial velocity(u) = 0

final velocity (v) = 72 km/hr or 20 m/s

time taken = 5 min. or 300 seconds

to find : acceleration and distance travelled

here we go

solution : for acceleration

by using 1st equation of motion

v = u+at

20 = 0+300a

300a = 20

a = 1/15 or 0.066 m/s²

solution : for distance travelled

by using 3rd equation of motion

v² = u²+2as

(20)² = 0+2*1/15 *s

400 = 2/15s

s = 400*15/2

s = 3000 meter


hope it helps :)

Answer 2

Answer:

a=1/15 and S=3 km

Explanation:

We have been given

u=0,v=72 km/h

t=5 min

=5×60

=300 sec

1.Finding acceleration

a=v-u/t

=20-0/300

=1/15 m/s^2

2.Finding Distance by using

2as=v^2-u^2

S=v^2/2a

=20/2×1/15

=3000 m

3000 m=3 km

•The acceleration of the train is 1/15 m/s^2 and the distance travelled is 3 km.