Question

A wildebeest calf is cruising at its top speed of v= 10 m/s when it passes over a sleeping cheetah. By the time the cheetah stands up and begins pursuit, the wildebeest is d= 7.0 ma head of the cheetah. If the cheetah is able to maintain a constant acceleration of a= 9.5 m/s^2until it catches the wild-beast, then how much time t must pass before the cheetah catches up to the wildebeast? (Hint: you will need to usethe quadratic formula.) For the limits check, investigate what happens to t as the initial separation distance approaches zero.

Answer 1

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

The given parameters;

• speed of the wildebeest calf, Vw = 10 m/s

• distance traveled by the calf before the cheetah stands up = 7 m

• constant acceleration of the cheetah, a = 9.5 m/s²

Let the speed of the cheetah = Vc

let the time the cheetah catches up with the wildebeest = t

$a = \frac{V_c}{t}$

$V_c = at$

Apply relative velocity formula to determine the time when the cheetah catches up with the widebeest;

Assuming the wildebeest and the cheetah are running in the same direction;

$(V_c - V_w) t = 7 \\\\(at - V_w) t = 7\\\\(9.5t - 10)t = 7\\\\9.5t^2 - 10t = 7\\\\9.5t^2 -10t-7 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 9.5 \ b= -10 \ , c =-7\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-10)\ \ +/- \ \ \sqrt{(-10)^2-4(9.5\times -7)} }{2(9.5)}\\\\t = \frac{10 \ \ +/- \ \ 19.13}{19} \\\\t = 1.53 \ s$

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero;

$(V_c - V_w) t = 0\\\\(at - V_w) t = 0\\\\(9.5t - 10)t = 0\\\\9.5t^2 - 10t = 0\\\\9.5t^2 = 10t\\\\9.5t = 10\\\\t = \frac{10}{9.5} = 1.05 \ s$

Thus, if the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

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