If its speed at the start of the 4th second is v, then t seconds later, the distance is s = vt + 1/2 at^2
So now you know that
s(1)-s(0) = 6
s(2)-s(1) = 7
That is,
v + 1/2 a = 6
(2v + 2a) - (v + 1/2 a) = 7
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2v+a = 12
2v+3a = 14
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a = 1
v = 11/2
A particle is travelling in a straight line with constant acceleration. It covers 6m in the 4th second and 7m in the 5th second,find its acceleration and the initial speed.
1 answer
1 answer