A zebra sees a lion 300 m to his right. The zebra starts running at 25 KpH straight ahead to get away. At this point the lion launches himself on an intercept course at the zebra at a speed of 31 KpH.
a. At what angle does the lion launch himself?
b. If both animals continue to travel straight, how long does it take the lion to catch the zebra?
clearly the angle is x (from directly ahead), where
sin(x) = 25/31
If it takes t hours, then
0.3^2 + (25t)^2 = (31t)^2
To find the angle at which the lion launches himself, we can use trigonometry. The given information is that the zebra sees the lion 300 meters to his right, so we can consider this as a right triangle. Let's label the right angle as C, the distance between the zebra and the lion as a (300 meters), and the distance the zebra runs as b.
Using the Pythagorean theorem, we can find the distance the zebra runs (b):
b^2 = 300^2 + X^2
b^2 = 300^2 + (25 * t)^2
b^2 = 90000 + 625t^2
Now, let's find the distance the lion runs (c):
c = 31 * t
Since c is the hypotenuse of the right triangle, we can use the law of cosines to find the angle at which the lion launches himself (angle A):
c^2 = a^2 + b^2 - 2ab * cos(A)
(31t)^2 = 300^2 + (90000 + 625t^2) - 2 * 300 * sqrt(90000 + 625t^2) * cos(A)
Simplifying the equation and isolating the cosine term:
(31t)^2 - 300^2 - 90000 - 625t^2 = - 600 * sqrt(90000 + 625t^2) * cos(A)
(961t^2 - 90000) - 625t^2 = - 600 * sqrt(90000 + 625t^2) * cos(A)
336t^2 - 90000 = - 600 * sqrt(90000 + 625t^2) * cos(A)
Dividing by 600:
0.56t^2 - 150 = - sqrt(90000 + 625t^2) * cos(A)
Squaring both sides to eliminate the square root:
0.3136t^4 - 168t^2 + 22500 = 90000 + 625t^2
Rearranging the equation:
0.3136t^4 - 168t^2 - 625t^2 + 67500 = 0
0.3136t^4 - 793t^2 + 67500 = 0
At this point, we have a fourth-degree polynomial equation that can be solved using numerical methods like Newton's method or graphing calculators. Unfortunately, finding an exact solution using algebraic methods is not practical.
To find the time it takes for the lion to catch the zebra, we equate the distances they traveled:
31t = 25t
6t = 300
t = 50 seconds
Therefore, it takes the lion 50 seconds to catch the zebra if both animals continue to travel straight.