Verified answer

Question:-

A bus moves away from rest at a bus stop with acceleration of 1 m/s². As the bus starts to move a man who is 4 m behind the stop with a constant speed after the bus. If he just manage to catch the bus find his speed?

Given:-

[tex] \implies[/tex] Let the man will catch the bus after time 't' since the distance from man position is same for both at time 't'

Solution:-

Distance covered by bus = Distance covered by man -−

[tex]\begin{gathered} = \frac{1}{2} at^{2} = vt - 48 \\ \\ \\ \frac{1}{2}^{(1)t2 }\end{gathered}[/tex]

[tex]\sf\bold\blue{Then,}[/tex]

[tex]\begin{gathered} = {t}^{2} - 20t + 56 = 0 \\ \\ (t - 12) \: (t - 8) = 0\end{gathered}[/tex]

[tex]t = 8s \: (manimum \: value)[/tex]

[tex]\rule{230pt}{2pt}[/tex]

Answer:-

So at t = 8s , the man catch the bus .

8s = 8 speeds

[tex]\rule{230pt}{2pt}[/tex]

Answer:

Distance covered by bus = Distance covered by man -−

\begin{gathered}\begin{gathered} = \frac{1}{2} at^{2} = vt - 48 \\ \\ \\ \frac{1}{2}^{(1)t2 }\end{gathered}\end{gathered}

=

2

1

at

2

=vt−48

2

1

(1)t2

[tex]Then,

\begin{gathered}\begin{gathered} = {t}^{2} - 20t + 56 = 0 \\ \\ (t - 12) \: (t - 8) = 0\end{gathered}\end{gathered}

=t

2

−20t+56=0

(t−12)(t−8)=0

[/tex]

So at t = 8s , the man catch the bus .

8s = 8 speeds

\rule{230pt}{2pt}