Question

a bus starts from rest with a constant acceleration of 5 m/sec Square. at the same time a car travelling at a constant velocity of 50m/sec overtakes the Bus and passes it. find at what distance will the bus overtake the car and how fast the bus would be travelling then

Answer 1

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

PLZ MARK IT AS BRAINLIEST ANSWER AND DROP A ♥

Answer 2

Answer:

100m/s

Explanation:

Bus: v= u +at

= 0+5t

=5t

s= ut+1/2 at²

=0t+1/2(5)t²

=5/2t²............(1)

v=s/t

Thus, s=vt

=50t...........(2)

From (1) and (2)

50t=5/2t²

Thus, t=20 s

Bus: a=v/t

v=at

=5*20

=100m/s