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  • A man is standing  behind the bus. Bus starts with <img alt="1 \mathrm{~m} / \ma

A man is standing 40 \mathrm{~m} behind the bus. Bus starts with 1 \mathrm{~m} / \mathrm{sec}^2 constant acceleration and also at the same instant the man starts moving with constant speed 9 \mathrm{~m} / \mathrm{s}. The time taken by man to catch the bus?

Option: 1

\mathrm{8\, \mathrm{sec}}


Option: 2

\mathrm{6 \, sec}


Option: 3

8 \, \mathrm{sec} \text { or } t=10 \, \mathrm{sec}


Option: 4

\mathrm{\text { None }}


Answers (1)

best_answer

let after time ' t ' man wile catch the bus.
for bus -
\mathrm{ \begin{aligned} & x=x_0+u t+\frac{1}{2} a t^2, \quad x=40+0(t)+\frac{1}{2}(1) t^2 \\ & x=40+\frac{t^2}{2}-(1) \end{aligned} }
for man - x=g t-(2)

from (1)  &  (2) - 

             \mathrm{\begin{aligned} & 40+\frac{t^2}{2}=9 t \\ & 80+t^2=18 t \\ & t^2-18 t+80=0 \end{aligned}}

Now, 

                \mathrm{\begin{aligned} & t^2-(10+8) t+80=0 \\ & t^2-10 t-8 t+80=0 \end{aligned}}

         \begin{aligned} & t(t-10)-8(t-10)=0 \\ & (t-10)(t-s)=0 \\ & t-10=0, t-8=0 \\ & t=10 \, \mathrm{sec}, t=8\, \mathrm{sec} \text { Ans } \end{aligned}

Posted by

Anam Khan

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