Question

A deer moving with uniform velocity 10m/s is being chased by a leopard, when will it catch the deer, if it was initially 10 m behind and acceleration of leopard is 1.56 m/s2, given that the leopard was initially at rest.
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Answer 1

The leopard will catch the deer after 13.7 seconds.

Explanation:

The deer is moving by uniform motion (=constant velocity), therefore we can write the equation of its position at time t as follows:

$x_d(t) = v_d t + x_0$

where

$v_d = 10 m/s$ is the velocity of the deer

t is the time

$x_0 = 10 m$ is the initial position of the deer when the leopard starts moving (in fact, the leopard starts from 10 metres behind)

Substituting the numbers,

$x_d(t) = 10+10t$ (1)

Instead, the leopard moves from rest at constant acceleration, so the equation of its position is

$x_l(t) = \frac{1}{2}at^2$

where

$a=1.56 m/s^2$ is its acceleration

Substituting,

$x_l(t) = \frac{1}{2}(1.56)t^2=0.78 t^2$ (2)

The leopard catches the  deer when they are at the same position, therefore

$x_d = x_l\\10+10t=0.78t^2$

And solving the equation,

$0.78t^2 -10t-10 = 0\\t=\frac{+10\pm \sqrt{(-10)^2-4(0.78)(-10)}}{2(0.78)}$

which gives two solutions:

t = -0.93 s

t = 13.7 s

And neglecting the negative solution since it has no physical meaning, we can say that the leopard catches the deer after 13.7 seconds.

Learn more about accelerated motion:

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