Question

A particle starting from rest move with constant acceleration if it takes 5sec to reach the 18km/hr find a average velocity b distance

Answer 1

initial velocity, u = 0
Final velocity, v = 18 km/h = 18 × 5/18 = 5 m/d
Time taken, t = 5 s

We know that v = u + at

5 = 0 + a×5

5 = 5a

a = 5/5 = 1 m/s^2

Distance, s = ut + 0.5at^2 = 0 + 0.5×1×5^2 = 12.5 m

Average velocity= 12.5/5 = 2.5 m/s

Answer 2

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initial velocity, u = 0

Final velocity, v = 18 km/h = 18 × 5/18 = 5 m/d

Time taken, t = 5 s

We know that v = u + at

5 = 0 + a×5

5 = 5a

a = 5/5 = 1 m/s^2

Distance, s = ut + 0.5at^2 = 0 + 0.5×1×5^2 = 12.5 m

Average velocity= 12.5/5 = 2.5 m/s