Asked by Maihul
A bus starts frm rest and moves with a uniform acceleration of (1m/secondsquare) for 5 minutes. calculate the velocity and distance?? plz explain and also give me the correct answer
Answers
Answered by
Anonymous
Bus starting from rest means Vi and Xi = 0
(where Vi is initial velocity and Xi is initial position).
5 mins = 300 seconds.
Vf = Vi + a*t (Vf is final velocity, a is acceleration, t is time).
Therefore,
Vf = 0 + 300*1 = 300 m/s.
For distance,
Xf = Xi + Vi*t + a*t^2*(0.5).
Xf = 0 + 0*t + (1)*(300)^2*0.5
Xf = 45000 m.
(where Vi is initial velocity and Xi is initial position).
5 mins = 300 seconds.
Vf = Vi + a*t (Vf is final velocity, a is acceleration, t is time).
Therefore,
Vf = 0 + 300*1 = 300 m/s.
For distance,
Xf = Xi + Vi*t + a*t^2*(0.5).
Xf = 0 + 0*t + (1)*(300)^2*0.5
Xf = 45000 m.
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