A tiger waits in hiding behind a bush. The tiger spots an impala running past at 10 m/s. At the instant the impala passes the tiger, the tiger accelerates from rest at 4 m/s² to catch the impala. The displacement of the tiger when it catches the impala is . Please answer.

the displacement is 50 meters.

Explanation:

We can assume that the initial position of both animals is the same, let's find the position equation for each one of them.

Impala:

We know that its velocity is 10m/s

then the velocity equation is

Vi(t) = 10m/s

For the position equation, we need to integrate over time to get:

Pi(t) = 10m/s*t + C

Where C is the constant of integration and represents the initial position of the impala.

For the Tiger, we know that the acceleration is 4m/s^2, then the acceleration equation is:

At(t) = 4m/s^2

The velocity equation can be found by integrating over time, we get:

Vt(t) = (4m/s^2)*t + V0

Where V0 is a constant of integration and represents the initial velocity of the tiger, we knew that the tiger was at rest, then the initial velocity is zero, and the velocity equation is:

V(t) = (4m/s^2)*t

For the position equation of the tiger we need to integrate over time again, and because both animals have the same initial position, the constant of integration will be equal to C.

Pt(t) = (1/2)*(4m/s^2)*t^2 + C

The tiger will catch the impala when the position of the tiger is the same as the position of the impala, then we need to solve:

Pi(t) = Pt(t)

10m/s*t + C =  (1/2)*(4m/s^2)*t^2 + C

We can remove C in both sides to get:

10m/s*t =  (1/2)*(4m/s^2)*t^2

Now we can divide both sides by t, to get

10m/s = (1/2)*(4m/s^2)*t

10m/s = (2m/s^2)*t

(10m/s)/(2m/s^2) = t = 5s

The tigger runs for 5 seconds before catching the Impala, knowing this we can find the displacement of the tiger, wich will be the difference between the final position of the tiger, Pt(5s) and the initial position of the tiger Pt(0s)

Displacement =  (1/2)*(4m/s^2)*(5s)^2 + C - ((1/2)*(4m/s^2)*(0s)^2 + C)

                       =  (1/2)*(4m/s^2)*(5s)^2 = 50m

The displacement of the tigger is 50 meters.

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