Question

A man is 48m behind a bus which is rest .The bus starts acceleration at the rate of 1m/second square at same time the man starts running with unifrm velocity of 10 m/s .what is the minimum time in which the man catches the bus exaplian it

Answer 1

$\boxed{ \bf \: Hello \: There}$

$\bf Given \: that$

Distance , s = 48 m

acceleration , a = 1m/s²

intial velocity , u = 0 m/s

Final velocity , v = 10m/s

time ,t = ?

$\bf \: By \: using \: first \: equation \: of \: motion \\ \\ = > \boxed{v = u + at = > 10 = 0 + 1 \times t = 10s}$

Hope this helps you ☺

Answer 2

Given,
Distance ( s ) = 48m,
Acceleration ( a ) = 1m/s square,
Initial velocity ( u ) = 0 m/s
Final velocity ( v ) = 10m/s,
So, time ( t ) = ?
As, we know that
s = ut + 1/2 at square
s = 0*t + 1/2*1*t*t
s = 0 + 1/2t square
48m = 1/2t square
t square = 48 * 2 = 96
t = √96 = 9.8sec = 10 sec ( approx).
Therefore, the minimum time in which the man catches the bus is 9.8 seconds.
Thanks!