Question

A man is standing 40 m behind the bus. Bus starts with 1m per second square constant acceleration and also at the same instant the man starts moving with constant speed 9mper sec. Find the time taken by man to catch the bus

Answer 1

INITIAL VELOCITY OF BUS = 0

FROM SECOND EQUATION OF UNIFORMLY ACCELERATED MOTION,

WE HAVE,

$s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} \times 1 \times {t}^{2} \\ s = \frac{ {t}^{2} }{2} ........(1)$

WE KNOW THAT THE DISTANCE COVERED BY MAN WILL BE EQUAL TO THAT OF DISTANCE COVERED BE BUS IN ORDER TO CATCH THE BUS.

SO,

DISTANCE (S) = 9 × T ____ (2)

EQUALISING (1) & (2)

$\frac{ {t}^{2} }{2} = 9 \times t \\ \frac{ {t}^{2} }{2 \times t} = 9 \\ \frac{t}{2} = 9 \\ t = 18 \: seconds$

TIME TAKEN BY MAN TO CATCH THE BUS IS 18 SECONDS.

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