A car is waiting at a traffic signal and when it turns green. the car starts ahead with constant acceleration of 3 meter per second square. At the same time, a bus travelling with a constant speed of 20 meters per second overtake and passes the car.
How far beyond its starting point will the car overtake the bus?
What will be the velocity of the car at that time?
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Let
t = time taken for overtaking
Condition to overtake:
Distance travelled by car
=Distance travelled by bus
[ut + 1/2at^2] = [ut + 1/2at^2]
[(0)t + 1/2(3)t^2] = [20t + 1/2(0)t^2]
1.5t^2 = 20t
1.5t = 20
t = 13.33s
(a)Distance beyond the starting point to overtake the bus:
s = 20t
=20 × 13.33
=266.6m
(b)Velocity of car while overtaking the bus:
v = u+at
=0 + 3×13.33
=39.99m/s
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t = time taken for overtaking
Condition to overtake:
Distance travelled by car
=Distance travelled by bus
[ut + 1/2at^2] = [ut + 1/2at^2]
[(0)t + 1/2(3)t^2] = [20t + 1/2(0)t^2]
1.5t^2 = 20t
1.5t = 20
t = 13.33s
(a)Distance beyond the starting point to overtake the bus:
s = 20t
=20 × 13.33
=266.6m
(b)Velocity of car while overtaking the bus:
v = u+at
=0 + 3×13.33
=39.99m/s
Is this answer helpful
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