IIT JEE , AIEEE Forum - Physics , Mechanics

s.mirchi

a bus starts frm rest with a constant acceleration of 5cm/s² . at the same time a car travelling with a constant velocity of 50cm/s overtakes and passes the bus.find-

a) at wht distance will the bus overtake the car?

b) hw fast will the bus b traveling then?

Aatish

let us consider for the bus:
u is initial velocity which is 0
a be the acceleration = 5 cm/sec
t be the time taken to cover x cm of distance where x is the distance in which bus overtakes the car.

During that period the car would also have covered same distance in the same time.

so for the bus.......

using, s = ut + (1/2)at²
we will get the relation b/w x and t as

2x = 5t²

also from car we will get x = 50 t

solving for x and t we get

x = 1000 cm

and then using v² - u² = 2as for bus we get velocity of bus at the instant of overtaking as v = 100 cm/sec

pls correct me if i am wrong somewhere................

iitkgp_bipin

At the instant of overtaking both have covered the same distance, so equate their distances.

BUS :

u=0 , a=5
s = ut + (1/2)at²
s = (1/2)(5)(t²)............(1)

CAR :
s = ut (since acceleration is zero)
s = 50t.............(2)

Substituting (2) in (1) :

50t = (5/2)(t²)
t = 20 sec

so, s = 50t = 1000 cm = 10 m

speed of bus at that instant :

v = u + at = 0 + (5)(20) = 100 cm/s = 1 m/s

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