Alion 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?

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.

i was thinking more like reflects green light, just not sure