JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2017
- Hard
- MCQ
A car is standing \(200\, m\) behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration \(2\,m/s^2\) and the car has acceleration \(4\, m/s^2\) . The car will catch up with the bus after a time of
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(10\sqrt {2}\, s\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Give,\,{u_c} = {u_B} = 0,{a_C} = 4m/{s^2},{a_B} = 2m/{s^2}\\
hence\,relative\,acceleration,\,{a_{CB}} = 2m/{\sec ^2}\\
Now,\,we\,know,\,s = ut + \frac{1}{2}a{t^2}\\
200 = \frac{1}{2} \times 2{t^2}\,\,\,\,u = 0\\
Hence,\,the\,car\,will\,catch\,up\,with\,the\,bus\,after\,time\,\\
t = 10\sqrt {2\,} {\rm{second}}
\end{array}\)
Give,\,{u_c} = {u_B} = 0,{a_C} = 4m/{s^2},{a_B} = 2m/{s^2}\\
hence\,relative\,acceleration,\,{a_{CB}} = 2m/{\sec ^2}\\
Now,\,we\,know,\,s = ut + \frac{1}{2}a{t^2}\\
200 = \frac{1}{2} \times 2{t^2}\,\,\,\,u = 0\\
Hence,\,the\,car\,will\,catch\,up\,with\,the\,bus\,after\,time\,\\
t = 10\sqrt {2\,} {\rm{second}}
\end{array}\)
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