What is the solution to get the acceleration of the lion in order to run the same distance as its prey?

Question

A lion is trying to catch an antelope. The antelope runs for 12.0s at a constant velocity of 36.0m/s. Starting from rest, what constant acceleration must the lion maintain in order to run the same distance as its prey runs at the same time?

Answer 1

Hi Azalel B.,

Unfortunately a crucial datum is missing here: what was the distance that the lion started behind the antelope? And, it really doesn't do to say they started from the same point, since then the lion would have just eaten the antelope before it trotted away. And, when you DO solve for d = 12 s * 36 m s^-1 using d=(1/2)at^2 and t=12 s, and then compare the lion's final speed (v=at), I bet you find that 74 m s^-1 is way faster than even a cheetah could manage. Just sayin. (70 mph max, with 60 mph ~ 88 ft/s^-1 -- you do the conversions)

Maybe the lion is .... "driving an Impala"?

Good math, poor biology ....

-- Cheers

Answer 2

Assuming they start from the same place:

antelope before the lion starts running:

vi = 36 m/s

t = 12.0 s

We can find how far the antelope has traveled by using

v = distance / time (used for constant velocities)

36 = d / 12

d = 432 meters

Now we can find the information for the lion, assuming it started from rest and must travel this same distance in the same time:

d = 432m

vi = 0

t = 12 seconds

We have d, vi, and t, and we are looking for acceleration a

use x = Vit + 1/2at^2 (x is the displacement)

432 = 0*12 + 1/2 * a * 12^2

432 = 72a

a = 6m/s^2

If an antelope runs 432m in 12 seconds at 36m/s constantly, a lion starting from rest would have to accelerate at a=6m/s^2 to accomplish the same distance in the same time.

(I agree this is not a good question. It would have been better if they had given the antelope a head start of 12 seconds and then asked you the acceleration needed for the lion to catch up within a certain distance or time.)