Question

A body starts from rest and moves with constant acceleration for t seconds. It travels a distance x1 in first half of the journey and distance x2 in the next half of the time,then. 

Answer 1

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

HI THERE

HERE IS UR ANSWER

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body stars from rest , there fore

u = 0 m/s

EQN used

s = u.t + 1/2 a.t^2

where ,

u is initial velocity

t is time

a is acceleration

HOPE HELPED

:-)