Question

When a deer was 48m from a leopard, the leopard starts chasing the deer and the deer immediately starts running away from the leopard with constant velocity. A leopard cannot run at high speed for a long time and has to slow down due to fatigue. If we assume that the Leopard starts with an initial speed of $30m/s$ and reduces its speed in equal steps of $5m/s$ after every $2\sec $ interval, At what minimum speed must the deer run to escape from the leopard?

Answer 1

Hint: If the deer wants to escape from the leopard then the deer has to be run until the leopard stops. First we calculate for what time the leopard can run and after that we calculate what is the maximum distance covered by the leopard in that time. Deer should cross the stopping point of leopard in that time then only deer can escape from leopard.

Complete step by step solution:
It is given in the question the speed of leopard decrease by $5m/s$ in every $2\sec $ so the deceleration of leopard is $\dfrac{5}{2}m/{s^2}$
The initial velocity of leopard is given $30m/s$
The acceleration of leopard is $ - \dfrac{5}{2}m/{s^2}$
The final velocity of leopard become zero when leopard come to rest
Here we can calculate the running time of leopard
By first Equation of motion
$ \Rightarrow v = u + at$
Put the given values
$ \Rightarrow 0 = 30 - \dfrac{5}{2}t$
So time of running of leopard
$ \Rightarrow t = \dfrac{{30 \times 2}}{5}$
$\therefore t = 12\sec $
Hence after $12$ second leopard will stop
Now we calculate in $12\sec $ how much the distance will be covered by the leopard
Third equation of motion
$ \Rightarrow {v^2} = {u^2} + 2as$
Apply given values
$ \Rightarrow 0 = {\left( {30} \right)^2} - 2 \times \dfrac{5}{2} \times s$
So the distance $s$covered by leopard in $12\sec $
$ \Rightarrow s = \dfrac{{900}}{5}$
$\therefore s = 180$ Metre
So we found the leopard can run only$180$ metre from it’s the starting point
If the deer wants to skip from leopard then the deer has to be Cross stopping point of leopard within$12$sec
It is given in question the leopard is behind by $84$ metre so the deer had to be run$\left( {180 - 84} \right) = 132$ metre in$12$ seconds
So we can calculate the velocity of deer
$ \Rightarrow v = \dfrac{{132}}{{12}}$
$ \Rightarrow v = 11m/s$

So the deer had to run at a minimum speed of $11m/s$.

Note: As we know the acceleration is defined as rate of change of velocity means how much the velocity changed in $1$ sec. In question it is given that the speed of leopard decrease in every $2$ sec by $5m/s$with the help of this statement we find de-acceleration of the leopard so change in velocity in one second is find as $\dfrac{5}{2}m/{s^2}$.