A cheetah spots a thomsons gazelle its preferred prey and...

A cheetah spots a Thomsons 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. a) Does the gazelle escape? b) On the same graph sketch the position as a function of time of the gazelle and the cheetah, assuming t = 0 is the time when the gazelle sees the cheetah.

Transcript

00:01 So on this question we have a cheetah.

00:03 This is a cheetah here.

00:04 This one, this black dots.

00:07 The spot a gazelle, but gazelle is the white dot.

00:11 And then the cheetah is moving at its...

00:15 It accelerates your stop speed at 10 meters of second and just as it gets your stop speed.

00:19 The gazelle spots it and starts running away, accelerating at 4 .6 mcm squared from rest.

00:27 So the cheetah was 1 70 meters away from the gazelle when it was.

00:31 Started running so we need to find out whether the chita will catch the gazelle so when the if the cheetah is to catch the gazelle don't reach a pascalar point at the same time so they have to reach the same this point at the same time and the distance for the cheetah to get to that point would be 170 plus the distance for the gazelle to get to that point that's called this so since they're going to reach this point at the same time we can equate their times.

01:15 So the time velocity is equal to distance of our time.

01:19 So time will be equal to distance by the bad velocity.

01:25 So the time for the cheetah be equal to 170 plus.

01:35 It's called this essence that so we don't confuse it with the other d.

01:44 Divided by the velocity of the cheetah which is good.

01:49 Now for b, for the gazelle, it's initial velocity, u will be zero met up a second.

02:00 So i'm going to use the equation as is equal to u3 plus half, 8t squared, and s is the distance here.

02:09 T is the time, use the initial velocity, this is the acceleration.

02:13 So substituting s is equal to zero, t, which will be zero, which we have times the acceleration 4 .6 times t squared so what we're going to do is find the distance and the time solving this simultaneously so i'm just going to put this in the calculator this equation one is equation two so when we solve this is to find that there's no real solution to this just squared simultaneous equations so all the solutions are imaginary and so the cheetah would not catch the gazelle so a the gazelle escapes no real solution for the next part once it gets the position on time graph so on the y axis we have the position and the x axis will have the time it's called this x for the position and t for the time and this x is going to be a method as unit time and seconds.

03:43 At the time, t is equal zero is when the cheetah sees the gazelle...

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