Question

A man is standing 40 m behind the bus. Bus starts with 1 m/s² constant acceleration and also at the same instant the man starts moving with constant speed 9 m/s. The minimum time taken by man to catch the bus will be?

Answer 1

here x∅=40m;

a=1m/s²;

Uman=9m/s;

x=0

➡let after time 't' man will catch bus,

for bus

x=x∅+ut+at²/2

x=40+0t+1t²/2

x=40+t²/2............(1)

for man

Uman=x/t

9=x/t

x=9t...............(2)

from equation (1) and (2)

40+t²/2=9t

80+t²/2=9t........(LCM)

80+t²=18t

t²-18t+80=0

t²-10t-8t+80=0

t(t-10)-8(t-10)=0

(t-8)(t-10)=0

t-8=0 t-10=0

t=8sec t=10sec

➡right answer is t=10sec (according to position-time graph)

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