Answer :
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:
[tex]x_d(t) = v_d t + x_0[/tex]
where
[tex]v_d = 10 m/s[/tex] is the velocity of the deer
t is the time
[tex]x_0 = 10 m[/tex] is the initial position of the deer when the leopard starts moving (in fact, the leopard starts from 10 metres behind)
Substituting the numbers,
[tex]x_d(t) = 10+10t[/tex] (1)
Instead, the leopard moves from rest at constant acceleration, so the equation of its position is
[tex]x_l(t) = \frac{1}{2}at^2[/tex]
where
[tex]a=1.56 m/s^2[/tex] is its acceleration
Substituting,
[tex]x_l(t) = \frac{1}{2}(1.56)t^2=0.78 t^2[/tex] (2)
The leopard catches the deer when they are at the same position, therefore
[tex]x_d = x_l\\10+10t=0.78t^2[/tex]
And solving the equation,
[tex]0.78t^2 -10t-10 = 0\\t=\frac{+10\pm \sqrt{(-10)^2-4(0.78)(-10)}}{2(0.78)}[/tex]
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.
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The speed of the leopard before it catches the deer is 11.36 m/s.
Speed of the leopard before it catches the deer
The speed of the leopard before it catches the deer is determined by applying the principle of relative velocity as shown below;
(Vl - Vd)t = d
where;
- Vl is the velocity of the leopard
- Vd is the velocity of the deer
- t is the time when the leopard catches the deer
- d is the distance between
(at - Vd)t = d
(1.56t - 10)t = 10
1.56t² - 10t = 10
1.56t² - 10t - 10 = 0
solve the quadratic equation using formula method;
t = 7.28 s
Velocity of the leopard
v = u + at
v = 0 + (1.56 x 7.28)
v = 11.36 m/s
Thus, the speed of the leopard before it catches the deer is 11.36 m/s.
Learn more about relative velocity here: https://brainly.com/question/17228388