a cheetah is 300 feet from its prey. it starts to sprint toward its prey at 90 feet per second. At the same time, the prey starts to sprint at 70 feet per second. when will the cheetah catch its prey?

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I will refer to the cheetah as C, and the prey as P. Okay, so the distance travelled by C at any point in time, t, is given by: S_c = 90t (This is found using distance = speed * time if that wasn't self-evident.) Similarly, the distance travelled by P is given by: _Sp = 70t Initially, the prey is 300 feet ahead of the cheetah, so to catch up, the cheetah needs to travel 300 metres more that the prey. That is: S_c = S_p + 300 Plugging in the values from our previous equations gives 90t = 70t  + 300 => 70t = 300 => t = 15 So the cheetah cathes its prey at 15s. We can check this by substituting t = 15 into our original equations: S_c = 90 * 15 = 1350 feet S_p = 70 * 15 = 1050 feet This is correct as the cheetah has travelled 300 feet more than the prey, accounting for the initial seperation distance.