1.

A bus starts from rest with a constant accelerationof 5 ms . At the same time a car travelling with aconstant velocity of 50 ms overtakes and passesthe bus. (i) Find at what distance will the busovertake the car ? (i) How fast will the bus betravelling then ? [Ans. () 1000 m (i) 100 ms1

Answer»

DISTANCE COVERED BY CAR WILL EQUAL TO THE DISTANCE COVERED BY BUS.

SO,

UT + 1/2 AT² = 50T

0×T+1/2×5×T² = 50T

5/2 T² = 50T

5/2 T = 50

T = 50×2/5

T = 20 SECONDS

DISTANCE = SPEED × TIME

DISTANCE = 50 × 20

DISTANCE = 1000 METRES

USING 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION,

V² = U² + 2AS

V² = 0² + 2×5×1000

V² = 10×1000

V = √10000

V = 100 M/S



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