Question

A train starting from rest moves with a uniform acceleration of 0.2 m/s^2 for 5 min. Calculate the speed acquired and the distance travelled in this time.

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Answer 1

Answer:

Explanation:

Given :-

Initial velocity of a train, u = 0

Time taken by the train, T = 5 min = 300 sec

Acceleration by the train, a = 0.2 m/s²

To Find :-

Final velocity, v = ???

Distance traveled, S = ???

Formula to be used :-

1st equation of motion i.e v = u + at

2nd equation of motion i.e S = ut + 1/2at²

Solution :-

1st Case :-

Final velocity of a train

v = u + at

= 0 + 0.2 × 300

= 60 m/s

Hence, speed acquired by the train is 60 m/s.

2nd Case :-

Distance traveled by the train

S = ut + 1/2at²

= 0 + 1/2 × 0.2 × (300)²

= 0.1 × 90000

= 9000 m

= 9 km

Hence, distance travelled by the train is 9 km.

Answer 2

Explanation:

u= 0m/s

v=?

a=0.2m/s^2

t=300 sec.

equation=1

v=u+at

now simply put the values

v=0+2/10x300

v=0+60

v=60m

to find distance we will use second equation

s=ut+1/2at^2

now put the values

s=0*300+1/2*2/10*300*300

s=0+9000

s=9000m/s

here s=distance

u=initial velocity

v=final velocity

t=time

a=acceleration

i hope it is correct and will help you.