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Question

For a body starting from rest and moving with uniform acceleration, the ratio of the distances covered for 1s,2s,3s,... is


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Solution

Step 1: Given

Initial velocity, u=0

Acceleration (a) is constant.

Step 2: Formulas used

From the three equations of motion, we know that,
s=ut+12at2
where s is the distance traveled, and t is the time taken

Step 3: Calculate distance covered for integral multiples of time

At t=1, distance covered is,
s1=0×1+12a×12=a2

At t=2,
s2=0×2+12a×22=22a2

At t=3,
s3=0×3+12a×32=32a2

Step 4: Calculate the ratio of distance covered

Thus, the ratio of the distance covered is,

s2s1=22a212a2=2212s3s2=32a222a2=3222snsn-1=tn2(tn-1)2
where subscript n denotes nth term of the same.

Therefore, the ratio of the distances covered for 1s,2s,3s,... is tn2tn-12.


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