A lion running at v = 11 m/s attacks a stationary impala lying directly ahead. The impala spies the lion when it is 11 m away. One-half second later the impala springs directly away with an acceleration of 8 m/s2. At what time after the impala catches sight of the lion does the lion catch the impala?

Let's model the displacement of the lion and impala as functions of time, L (t) and I (t) respectively.

We will model their motion from the perspective of where the lion was initially. This means the displacement of the lion is initially 0m. The impala has some amount of a head start, and thus its initial displacement is some positive number.

The lion is 11m from the impala. The impala takes 0.5 seconds to react to the lion and start running. This time delay means the lion is even closer to the impala when it starts running. To calculate the impala's head start, we will calculate:

x = vΔt

x is the head start, v is the lion's velocity, and Δt is the time delay.

Given values:

v = 11m/s

Δt = 0.5s

x = 11 (0.5) = 5.5m, thus the impala has a 5.5m head start.

It's not terribly difficult to mathematically model the lion's motion. It chases the impala at a constant velocity, therefore its displacement L (t) will be a linear function, ie a straight line with a slope of 11m/s:

L (t) = 11t

The impala moves under constant acceleration, so to model its motion we will use this kinematics equation:

I (t) = X + Vt + 0.5At²

t is time, X is the initial displacement, V is the initial velocity, and A is the acceleration.

Given values:

X = 5.5m (the head start)

V = 0m/s (the impala starts running from rest)

A = 8m/s²

Plug these values in to get I (t):

I (t) = 4t² + 5.5

What we need is to find when the lion catches the impala, ie calculate a time t for which L (t) = I (t), where the lion's displacement equals the impala's displacement. Simply set the functions equal to each other and solve for t:

L (t) = I (t)

11t = 4t² + 5.5

4t² - 11t + 5.5 = 0

You can use the quadratic formula to find the roots of the left-hand side, and therefore the values of t:

t = 0.657, 2.093

We have two values for t, but we are only looking for the first value of time where the lion meets up with the impala.

t = 0.657s

1.16 seconds

Explanation:

Let's say that t is the time since the impala catches sight of the lion.

The position of the lion is:

x = x₀ + v₀ t + ½ at²

x = (0) + (11) t + ½ (0) t²

x = 11t

The position of the impala is:

x = x₀ + v₀ t + ½ at²

x = (11) + (0) (t-0.5) + ½ (8) (t-0.5) ²

x = 11 + 4 (t-0.5) ²

Now we set the positions equal:

11t = 11 + 4 (t-0.5) ²

11t = 11 + 4 (t² - t + 0.25)

11t = 11 + 4t² - 4t + 1

0 = 4t² - 15t + 12

Solving with quadratic formula:

t = (15 ± √33) / 8

t ≈ 1.16, 2.59

We want the lesser time, so t = 1.16.

Notice that how you define t is very important. If you define it as the time since the impala starts running, you get an answer of 0.657, which is fine, but you have to remember that the problem asks for the time after the impala catches sight of the lion. So you need to add the half second delay to that. 0.66 + 0.50 = 1.16.