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 rate with acceleration `beta` starts from rest. For what value of `beta` will the overtake the rat?. (in `m s^(-2)`).

A cat, on seeing a rat at a distance `d=5 m`, starts velocity `u=5 m s

1.	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 rate with acceleration `beta` starts from rest. For what value of `beta` will the overtake the rat?. (in `m s^(-2)`).
Answer» Correct Answer - 5 For rat `S=(1)/(2) betat^(2)` (i) For cat `S =d =ut +(1)/(2) alpha t^(2) 0` Putting the value os `S` from Eq. (i) in Eq. (ii), `(a-b) t^(2) + 2 ut -2d =0` ` t=(2u+-sqrt( 4u^(2)-8 d (beta-alpha)))/(2(beta -alpha)` For `t` to be real, `(u^(2))/(2d) le (beta-alpha) becuase vbeta =alpha (u^(2))/(2d)` Substituting , a, d and `u` we get `beta =2.5 +(5^(2^(.)))/(2xx5) =2.5 +2 .5 =5 m s^(-2)`.

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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 rate with acceleration `beta` starts from rest. For what value of `beta` will the overtake the rat?. (in `m s^(-2)`).

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