Asked by Francis
A car A is 300m behind a car B and both are travelling in the same direction at 25m/s. Both cars begin at the same moment to accelerate at steady rate, but A has a greater acceleration and passes B in 20sec by which time B is doing 27•5m/s. Neglecting the length of the cars, calculate the acceleration of A, it's speed at the moment of passing B and the distance it has travelled in the 20sec.
Answers
Answered by
Henry
V = Vo + a*t = 27.5, 25 + a*20 = 27.5, 20a = 2.5, a = 0.125 m/s^2. = Acceleration of car B.
Vo*t + 0.5a*t^2 = Vo*t + o.5a*t^2 + 300,
Vo*t-Vo*t + 0.5a*t^2 = 0.5*0.125*20^2 + 300,
0.5a*20^2 = 25 + 300 = 325,
200a = 325, a = 1.625 = Acceleration of car A.
Va = Vo + a*t = 25 + 1.625*20 = 57.5 m/s. = Speed of car A at the moment of passing B.
d = 25*20 + 1.625*20^2 = 1150 m.
=
Vo*t + 0.5a*t^2 = Vo*t + o.5a*t^2 + 300,
Vo*t-Vo*t + 0.5a*t^2 = 0.5*0.125*20^2 + 300,
0.5a*20^2 = 25 + 300 = 325,
200a = 325, a = 1.625 = Acceleration of car A.
Va = Vo + a*t = 25 + 1.625*20 = 57.5 m/s. = Speed of car A at the moment of passing B.
d = 25*20 + 1.625*20^2 = 1150 m.
=
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