Question

An object start from rest moving with constant acceleration of 1 m/s. The distance travel between 2sec to 3 sec is​

Answer 1

Explanation:

Let us, take an example of a free fall.

I'm trying to explain this by an example cause it's a multiple choice question and also that it's based on the concept and not any numerical solving.

Alright !!

Let's suppose that a body is falling towards the earth vertically downwards.

So, it's acceleration will be the acceleration due to gravity that is 9.8 m/s^2.

And let's suppose that is starts from rest and falls for 2 seconds.

So,

let's calculate the distance travelled by it in these time intervals that is 0-1 sec and then 1-2 sec.

For, distance travelled in 1 sec

x1 = ut1 + 1/2 * g * (t1)^2

= 0*1 + 1/2*9.8*1^2

= 4.9 m

Now,

distance travelled in 2 seconds.

distance = ut2 + 1/2*g*(t2)^2

= 0*2 + 1/2*9.8*(2)^2

= 19.6 m

So, the distance travelled in the time interval 1-2 sec will be 19.6-4.9 = 14.7 m

Now this distance 14.7 is X2.

Thus, we observe that 14.7 = 3 * 4.9

That is

X2 = 3 x1.

Hence the correct answer is (c).

Hope this helps you !!

Answer 2

In 2 sec dist. travelled = 4m