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A cat, on seeing a rat at a distance d=5...

A cat, on seeing a rat at a distance `d=5 m`, starts velocity `u=5 m s^(-1)` and moves with acceleration `alpha =2.5 m s^(-2)` in order to catch it, while the rat with acceleration `beta` starts from rest. For what value of `beta` will the cat overtake the rat?. `(m s^(-2))`.

Text Solution

Verified by Experts

The correct Answer is:
5
For rat `S=(1)/(2)betat^(2)` .(i)
For cat `S=d=ut+(1)/(2)alphat^(2)` …(ii)
puttin the value of S from Eq. (i) in Eq (ii),
`(a-b)t^(2)+2ut-2d=0`
`t=(2u+-sqrt(4u^(2)-8d(beta-alpha)))/(2(beta-alpha))`
For t to be real `(u^(2))/(2d)ge(beta-alpha) :. beta=alpha+(u^(2))/(2d)`
substituting a d and u we get
`beta=2.5+(5^(2))/(2xx5)=2.5+2.5=5ms^(-2)`
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