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 = 7.0m ahead of the cheetah. If the cheetah is able to maintain a constant acceleration of a = 9.5 m/s2 until it catches the wilde-beast, then how much time t must pass before the cheetah catches up to the wildebeest? (Hint: you will need to use the quadratic formula.) For the limits check, investigate what happens to t as the initial separation distance d approaches zero.

a. Representation.

b. Algebra Work.

c. Units Check.

d. Limits Check.

e. Define the Physics Concept.

Answer 1

For wilderbeast d = im v = 10 m/s Let distance Covered by wilderbeast before being caught - 2 vat . 2 = = lot For cheetah U=O a=9.5 mis distance = d ta = 7+ lot = 2 0 + using equation of motion ut & tat² 7 tot as t² = 7+lot 4.75 +²-lot - 7=0 On solving quadratic equ, we get: t = 2.66 seconds Time bossed 12.66 seconds.