Question

A train starting from rest move with a uniform acceleration of 0.2m/s2 for 5 min calculate the speed acquired and the distance travelled in this time?

Answer 1

hope its help you ......

Answer 2

Heya!!

Here is your answer : -

- Given,

Train starts from rest ,

Aquires an acceleration of 0.2m/s^2 for,

Time = 5 min

The values which we get from the above data :

u = 0

a = 0.2m/s^2

t = 5 min = 5 × 60 = 300 s

- To find ,

v = ?

S = ?

- Main solution :

First let's found final velocity by using 1st equation of motion i.e v = u + at

- We know,

u = 0

a = 0.2 m/s^2

t = 300 sec

- Now substituting the values :

v = 0 + (0.2)(300)

v = (0.2)(300)

v = 60 m/s ( Ans )

- Now we have to find distance travelled. We will find it using 2nd equation of motion i.e s = ut + $\frac{1at^2}{2}$

- We know ,

u = 0

t = 300 sec

a = 0.2 m/s^2

- Substituting the values :-

S = (0)(300)+ $\frac{1(0.2)(300)(300)}{2}$

S = 0 + 9000

S = 9000 m ( Ans )


=>> So, the distance travelled by the train is 9000 m and velocity ( speed ) acquired by it is 60m/sec.

Hope it helps you.