A man travelling in a car with maximum constant speed of 20 ms − 1 watches his friend, who starts off at a distance of 100 m ahead on a motor cycle with constant acceleration ' a '. The maxi

Question

A man travelling in a car with maximum constant speed of $20 ms^{- 1}$ watches his friend, who starts off at a distance of 100 m ahead on a motor cycle with constant acceleration ' a '. The maximum value of ' a ' for which the man in the car can reach his friend will be-

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Answer:

A

Solution:

$v = ks \Rightarrow \frac{dv}{dt} = k \frac{ds}{dt} \Rightarrow a = kv = k^{2} s$

TestHub · Accepted Answer

v = ks ⇒ dt dv ​ = k dt ds ​ ⇒ a = kv = k 2 s

Physics - Kinematics Question with Solution | TestHub

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