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JEE Mains · Physics · STD 11 - 2. motion in straight line
A scooter accelerates from rest for time \(t_{1}\) at constant rate \(a _{1}\) and then retards at constant rate \(a _{2}\) for time \(t _{2}\) and comes to rest. The correct value of \(\frac{t_{1}}{t_{2}}\) will be ..... .
- A \(\frac{a_{1}+a_{2}}{a_{2}}\)
- B \(\frac{ a _{2}}{ a _{1}}\)
- C \(\frac{ a _{1}}{ a _{2}}\)
- D \(\frac{a_{1}+a_{2}}{a_{1}}\)
Answer & Solution
Correct Answer
(B) \(\frac{ a _{2}}{ a _{1}}\)
Step-by-step Solution
Detailed explanation
Draw vt curve \(\tan \theta_{1}=a_{1}=\frac{v_{\max }}{t_{1}}\) And \(\tan \theta_{2}=a_{2}=\frac{v_{\max }}{t_{2}}\) \(\div\) above \(\frac{t_{1}}{t_{2}}=\frac{a_{2}}{a_{1}}\)
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