Question

A man is standing 40 m behind the bus. The bus starts with 1 \[m/{{s}^{2}}\] constant acceleration and also at the same instant the man starts moving with a constant speed of 9 m/s. Find the time taken by the man to catch the bus.
A. 6s
B. 10s
C. 4s
D. 13s

Answer 1

Hint: In this problem the man and the bus both are in motion. The man is moving with the constant speed which means its acceleration is zero while the busing is moving with a constant acceleration.

Complete step by step answer:
Distance between bus and man initially, s= 40m
Let after covering a distance X the man caught the bus and Let after time 't' man will catch the bus.
Total distance covered by the man = (X+40) m
Total distance covered by the bus = X m
For Man:
Using distance \[=speed\times time\]
X+40= 9t
X=9t-40------------(1)
For The Bus:
Initially at rest so u=0
a= 1\[m/{{s}^{2}}\]
distance=X m
using \[s=ut+\dfrac{a{{t}^{2}}}{2}\]
\[s=\dfrac{a{{t}^{2}}}{2}\]
\[X=0.5{{t}^{2}}\]-----------(2)
Eq(2) – Eq(1) we get \[0.5{{t}^{2}}-9t+40\]=0
Multiplying by 2 on both sides we get \[{{t}^{2}}-18t+80=0\]
Solving this quadratic equation,
\[\begin{align}
  & {{t}^{2}}-(10+8)t+80=0 \\
 & {{t}^{2}}-10t-8t+80=0 \\
 & t\{t-10\}-8\{t-10\}=0 \\
\end{align}\]

(t-8)(t-10)=0
Therefore, either t=8 or t=10
But in the options, we are only provided with t=10s

So, the correct answer is “Option B”.

Note:
While doing such kind of problems we have to keep in mind the number of variables which needs to be introduced and what are the given quantities. Also, we have to specifically keep in our mind whether we have to use equations of constant motion or it involves the use of acceleration.