JEE Mains · Physics · STD 11 - 2. motion in straight line
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
- A \(\sqrt {110}\, s\)
- B \(\sqrt {120}\, s\)
- C \(10\sqrt {2}\, s\)
- D \(15\,s\)
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\\…
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