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anchit saini

not as straightforward as it seems at first glance

Lets consider the case of one turning point only (all other cases can be shown equivalent to it )->

In this also , lets make 2 cases

Case 1 ->

In this case, lets consider the most symmetric case , ie when it turns at t=1/2 sec and it has same acceleration throughout

At t=1/2

s1=at^2 /2 =a/8

v1=at=a/2

t=1/2 to t=1

v2= 0 = u -At =a/2 - At =a/2- A/2

implies a = A

s2=ut - at^2/2 =a/4 - a/8 =a/8

but s1 + s2 = 1 = a/4

implies a = 4 m/s^2

the velocity time graph in this case is as shown in graph 1

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Case 2- >

Now , suppose instead of t=1/2 , the turning point comes at some other time , then it is obvious that if acceleration is less than 4 for interval before turning point , acceleration would be greater than 4 for for interval after turning point and vice versa

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Now , we have to proove that instead of 1 turning point suppose there are infinitely many turning points yet the above condition holds true .

For it , consider the graph 2 in which there are many turns .

Now we can consider any point P , and then we can see the the part before P will correspond to a certain average acceleration and part after P will correspond to a certain average acceleration , and hence the corresponding graph would come to be of either case 1 or case 2.

Now suppose , before P ,the average acceleration is greater than 4 , then since there are many turning points its obvious that to maintain average acceleration of more than 4 , at some points instantaneous accelerations would have to be greater than 4 .Same would be true for interval after P , if it has average acceleration greater than 4.

Hence , option a and c are correct

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