Question

A cheetah spots a thomson's gazelle, its preferred prey, and leaps into action, quickly accelerating to its top speed of 30 m/s, the highest of any land animal. however, a cheetah can maintain this extreme speed for only 15 s before having to let up. the cheetah is 170 m from the gazelle as it reaches top speed, and the gazelle sees the cheetah at just this instant. with negligible reaction time, the gazelle heads directly away from the cheetah, accelerating at 4.6 m/s2 for 5.0 s, then running at constant speed. does the gazelle escape? if so, by what distance is the gazelle in front when the cheetah gives up?

Answer 1

The gazelle escapes the cheetah by 7.5 meters. This is calculated by determining the distance the cheetah and gazelle each cover in the given time. The gazelle's distance includes both the acceleration phase and the constant speed phase.

Step-by-step explanation:

To determine if the gazelle escapes from the cheetah, we need to calculate the distance both the gazelle and the cheetah cover separately within the same time frame. The cheetah can maintain its top speed of 30 m/s for only 15 seconds. Therefore, the total distance covered by the cheetah while it's at top speed is given by:

Distance covered by the cheetah = speed × time = 30 m/s × 15 s = 450 m

The gazelle starts accelerating at 4.6 m/s2 for 5.0 seconds. The distance covered by the gazelle during acceleration can be calculated using the equation:

Distance = 0.5 × acceleration × time2 = 0.5 × 4.6 m/s2 × (5.0 s)2 = 57.5 m

After 5 seconds of acceleration, the gazelle will be running at a constant speed, which we can find using the formula:

Final velocity = initial velocity + (acceleration × time) = 0 + (4.6 m/s2 × 5.0 s) = 23 m/s

For the remaining 10 seconds (since the cheetah runs at top speed for 15 seconds and the gazelle has already spent 5 seconds accelerating), the gazelle travels at this constant speed, covering:

Distance at constant speed = speed × time = 23 m/s × 10 s = 230 m

The total distance covered by the gazelle is the sum of the distance covered during acceleration and the distance covered at constant speed:

Total distance covered by gazelle = 57.5 m + 230 m = 287.5 m

Now we need to add the initial distance between the gazelle and the cheetah to the distance covered by the gazelle, to find out how far the gazelle is when the cheetah stops:

Total distance from cheetah = initial distance + distance covered by gazelle = 170 m + 287.5 m = 457.5 m

Since the cheetah covers only 450 m and the gazelle is 457.5 m away from the cheetah's starting point, the gazelle escapes, and the distance by which it's in front when the cheetah gives up is:

Escape distance = total distance from cheetah - distance covered by cheetah = 457.5 m - 450 m = 7.5 m

Therefore, the gazelle escapes the cheetah by 7.5 meters.

Answer 2

We can find the gazelle's constant speed. v = a t = (4.6 m/s^2) (5.0 s) v = 23 m/s Note that the cheetah gains on the gazelle the whole time they are both running. We can find the distance the cheetah can run before it must stop. d = v t = (30 m/s) (15 s) = 450 m To escape, the gazelle has 15 seconds to travel 450 m -170 m which is 280 meters. We can find the distance x_1 the gazelle travels during the 5.0 second acceleration period. x_1 = (1/2) a t^2 x_1 = (1/2) (4.6 m/s^2) (5.0 s)^2 x_1 = 57.5 m We can find the distance x_2 the gazelle could run in the next 10 seconds. x_2 = v t = (23 m/s) (10 s) x_2 = 230 m The total distance the gazelle can travel in 15 seconds is 230 m + 57.5 m which is 287.5 meters. Since the gazelle only needed to run 280 meters to escape, the gazelle is able to escape from the cheetah with 7.5 meters to spare.