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

✺ Given ✺

✏ A deer escapes from a leopard and runs away in a straight line with constant velocity.

✏ The leopard again starts chasing the deer when the deer has run a distance of 64 m.

✏ The leopard starts with the initial speed of 30 m/s but due to fatigue it has to reduce its speed and it does so by reducing speed uniformly by 4 m/s every 2 second.

✺ To Find ✺

The speed the deer should run to save it's life​​.

✺ Solution ✺

✍ We must know before solving :-

Final velocity of leopard should be less than the velocity of the deer.

Initial velocity of leopard = 30 m/s

As it reduces 4 m/s every 2 seconds, so it is a deceleration.

Deceleration, a = -2 m/s²

Calculating the final velocity/speed of the deer of leopard :-

$\rm{ v=u+at}\\\\\rm{\dashrightarrow v=30+(-2)t}\\\\\rm{\dashrightarrow v=30-2t}$

✍ Now, when the leopard tries to catch the deer, that time, the final velocity of leopard = initial velocity of the deer.

✍ Both of them will cover the same distance, but the speed of the deer must be more, for which it runs ahead of the leopard.

Distance covered by the leopard :-

$\rm{(30-2t)^2=(30)^2+2(-2)s}\\\\\rm{\dashrightarrow 900+4t^2-120t=900-4s}\\\\\rm{\dashrightarrow 4t^2-120t+4s=0}\\\\\rm{\dashrightarrow 4s=120t-4t^2}\\\\\em{\dashrightarrow s=30t-t^2}$

As the travelled the same distance,

Distance covered by the deer :-

$\rm{\dashrightarrow s=30t-t^2=(30-2t)t+64}\\\\\rm{\dashrightarrow 30t-t^2=64+30t-2t^2}\\\\\rm{\dashrightarrow 30t-t^2+2t^2-30t=64}\\\\\rm{\dashrightarrow t^2=64}\\\\\rm{\dashrightarrow t=8\ secs}$

Velocity of the leopard = initial speed of the deer :-

$\rm{\dashrightarrow v=30-2(t)}\\\\\rm{\dashrightarrow v=30-2(8)}\\\\\rm{\dashrightarrow v=30-16}\\\\\rm{\dashrightarrow v=14\ m/s}$

Hence, the deer must have a speed more than 14 m/s to save his life.

Answer 2

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⏺ Required Answer:

✏GiveN:

• Initial speed of leopard = 30 m/s
• Distance between leopard and deer = 64 m
• Deceleration of leopard = 4/2 = 2m/s²
• Deer runs with constant velocity

✏To FinD:

• At what speed dear should run to save it's life.....?

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⏺ How to solve?

For solving this question, visualization is more important than the solving. After that, we can use the equations of motion to solve the question. The equations of motion:

$\large{ \rm{ \circ \: {v = u + at.........(1)}}} \\ \\ \large{ \rm{ \circ \: s = ut + \frac{1}{2} a {t}^{2}...........(2) }} \\ \\ \large{ \rm{ \circ \: { {v}^{2} = {u}^{2} + 2as..........(3)}}}$

Here, in this question we will use 1st and 2nd equation of motion.

Also we need the speed, distance and time relation i.e.

$\large{\circ\:{\rm{Distance = Speed \times Time}}}$

So, let's solve the question.

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⏺ Solution:

❇ The speed of the deer should be a bot greater than the speed of leopard at the exact moment, when they meet, so that the leopard won't catch the deer.

Let the time after they meet = t

And, velocity after time t will be

Applying first equation of motion, $\large{ \rm{ \longrightarrow \: v = u + at}} \\ \\ \large{ \rm{ \longrightarrow \: v_l = 30 + ( - 2)t}} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{v_l = 30 - 2t}}}}$

❇ The velocity of the dear should be bit greater than that of leopard, So

$\large{\rm{\longrightarrow \: v_d \geq 30 - 2t}}$

❇ At time t,

Dis. covered by leopard = Dis. covered by deer

So, let's find the distance covered by leopard and deer

Applying 2nd equation of motion,

$\large{ \rm{ \underline{ \red{tiger}}}} \\ \\ \large{ \rm{ \longrightarrow \: s_l = 30t + \frac{1}{2} \times (2) \times {t}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{s_l = 30t - {t}^{2} }}}} \\ \\ \large{ \underline{ \red{ \rm{deer}}}} \\ \\ \large{ \rm{ \longrightarrow \: s_d = 64 + (v_d \times t)}}$

❇ According to question:

$\large{ \rm{ \longrightarrow \: 30t - {t}^{2} = 64 + (v_d \times t)}} \\ \\ \large{ \rm{ \longrightarrow \: 30t - {t}^{2} = 64 + (30 - 2t)t}} \\ \\ \large{ \rm{ \longrightarrow \: 30t - {t}^{2} = 64 + 30t - 2 {t}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: {t}^{2} = 64}} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{t = 8s}}}}$

❇ So, From here velocity of leopard,

$\large{ \rm{ \longrightarrow \: v_l = 30 - 2(8) \: m {s}^{ - 1} }} \\ \\ \large{ \rm{ \longrightarrow \: v_l = 30 - 16 \: m {s}^{ - 1} }} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{v_l = 14 \: m {s}^{ - 1} }}}}$

❇ So, the minimum velocity needed by the dear to escape the tiger is $\geq$ 14 m/s

$\large{ \therefore{ \underline{ \underline{ \red{ \rm{Hence \: solved \: \dag}}}}}}$

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