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A body starting from rest is moving under a constant acceleration up to 20 sec. If it moves S1 distance in the first 10 sec, and S2 distance in the next 10 sec then S2 will be equal to:
a) S1
b) 2S1
c) 3S1
d) 4S1

Answer
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Hint: The body is moving with constant acceleration. Hence the speed of the body will keep on changing as time progresses. Hence using Newton’s second kinematic equation for the two different sections of the total distance, we will be able to determine the distance covered in the second section of the journey.
Formula used:
S=ut+12at2

Complete answer:
In the question it is given that the body moves S1 distance in the first 10 sec of the total journey. Let us say that the body is moving with acceleration ‘a’ and with initial velocity ‘u’. Then the distance (S) covered by the body in time ‘t’ from Newton’s second kinematic equation is given by,
S=ut+12at2
The body is initially at rest. Therefore its initial velocity is zero. Hence S1 is equal to,
S1=(0)t+12a(10)2S1=12a100∴⇒S1=50a....(1)
Let us say the same body has velocity ‘v’ at time t. hence from Newton’s first kinematic equation we get,
vu=at
For the next time interval of 10 sec, the body will have some initial velocity which is the final velocity of when it covered distance S1 . The initial velocity ‘x’ of the body hence is,
vu=atx0=a(10)x=10a
Hence from Newton’s second kinematic equation The distance S2 covered is,
S2=xt+12at2S2=10a(10)+12a(10)2S2=100a+50aS2=150aa=S150S2=150×S150S2=3S1

Therefore the correct answer of the above question is option c.

Note:
It is to be noted the value of ‘a’ can be obtained from equation 1. The body is accelerating with constant acceleration. If the body moves with random acceleration i.e. it changes with time, then Newton’s equations do not hold valid.
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