Question

a deer escapes from a leopard and runs away ina straight line with constant Velocity. the leopard again starts chasing the deer when the deer has run a distance of 64m. the leopard starts with the initial speed of 30m/s but due to fatigue it has to reduce its speed and it does so by reducing speed uniformly by 4m/s every 2 second.
at what speed should deer run to save it's life​

Answer 1

Answer:

The speed should deer run to save his life​ is 14 $ms^{-1}$

Explanation:

Deer is moving with constant velocity and leopard starts running with

30 $ms^{-1}$ speed and his speed decreases by 4 $ms^{-1}$ in 2 seconds.

⇒ Leopard's deacceleration = $\frac{Change in velocity}{time taken}$ = $\frac{4}{2}$ = -2 $ms^{-2}$

Speed of Leopard by linear motion equation:

→ $v$ = $u$ + $at$

→ $v$ = $(30) + (-2)t$

→ $v$ = $30 - 2t$ -------------------------------------- eq(1)

So, Leopard's sped will keep decreasing and at a time speed of both will be same, when Leopard will be about to catch the dear, but at the next moment his speed will decrease due to deacceleration and distance between them will continue to increase.

At a time,

⇒ Speed of leopard = Speed of deer

Both will cover same distance, but speed of deer must be more, so that he runs ahead of the leopard.

→ Applying $v^{2} - u^{2} = 2as$ equation for leopard to find distance covered using value of eq(1):

→ $(30 - 2t)^{2} - 30^{2} = 2(-2) s$

→ $900 + 4t^{2} - 120t - 900 = (-4)s$

→ $4s = 120t - 4t^{2}$

→ $s = 30t - t^{2}$ --------------------------- eq(2)

Both will cover same distance and distance covered

by deer is $(30 - 2t)t + 64$

→ So, $30t - t^{2}$ = $(30 - 2t)t + 64$

→ $t^{2} = 64$

→ $t = 8$ sec

Since both will have the same speed

putting value of t in eq(1)

⇒ $v = 30 - (2)(8)$

⇒$v = 14$ $ms^{-1}$