A bus starts from rest with a constant acceleration of 5m/s². Al the same time a car travelling with a constant velocity of 50m/s overtakes and passes the bus.
(i)Find at what distance will the bus overtake the car?
(ii)How fast will the bus be travelling then?
Class 11 motion in a straight line
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Let after time t the bus will over the car.
let s be the distance travelled by the bus to overtake the car.
For bus : As a = 5 m/s^2 ; u = 0 m/s
so, s = u t + 1/2 a t^2
or, s = 0 + 1/2 ×5 × t^2
so, s = 5/2 t^2
Also, the distance travelled by the car in the same time t with v = 50 m/s
d = v t = 50 × t
As both s and d is the same distance
so, s = d
or, 5/2 t^2 = 50 t
or, t^2 - 20 t = 0
or, t ( t - 20 ) = 0
so, t = 20 s ; { 0 s neglected }
a) hence after s or d distance the bus will overtake the car
so, d = 50 × t = 50 × 20 = 1000 m
b) also v = u + a t
so, velocity of bus after 20 s
v = 0 + 5 × 20 = 100 m/s
the bus will be travelling then at a speed of
100 m/s.
let s be the distance travelled by the bus to overtake the car.
For bus : As a = 5 m/s^2 ; u = 0 m/s
so, s = u t + 1/2 a t^2
or, s = 0 + 1/2 ×5 × t^2
so, s = 5/2 t^2
Also, the distance travelled by the car in the same time t with v = 50 m/s
d = v t = 50 × t
As both s and d is the same distance
so, s = d
or, 5/2 t^2 = 50 t
or, t^2 - 20 t = 0
or, t ( t - 20 ) = 0
so, t = 20 s ; { 0 s neglected }
a) hence after s or d distance the bus will overtake the car
so, d = 50 × t = 50 × 20 = 1000 m
b) also v = u + a t
so, velocity of bus after 20 s
v = 0 + 5 × 20 = 100 m/s
the bus will be travelling then at a speed of
100 m/s.
silenteye100:
thankyou
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