AP EAMCET · PHYSICS · Motion In One Dimension
A student is at a distance \(16 \mathrm{~m}\) from a bus when the bus begins to move with a constant acceleration of \(9 \mathrm{~m} \mathrm{~s}^{-2}\). The minimum velocity with which the student should run towards the bus so as the catch it is \(\alpha \sqrt{2} \mathrm{~ms}^{-1}\). The value of \(\alpha\) is
- A \(10\)
- B \(12\)
- C \(15\)
- D \(20\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
Let \(v\) be the minimum velocity of student so,that he could catch the bus If student catch the bus in time \(t\), then distance travelled by student in time \(t\) \(=16+\) distance travelled by bus in time \(t\).…
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