Question #a2cf2

1 Answer
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