Question #a2cf2

1 Answer
Jane
Jan 13, 2018

Because position increases by square of t (s prop t^2) whereas velocity increases by t(v prop t) for constant value of acceleration

Explanation:

Ref.
S= (1/2) a t^2(all symbols are bearing their conventional meaning)
V= at
Hence increase in amount of s will be more than v for increasing values of t

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