(5%) Problem 17: The world\'s fastest land animal, the cheetah, can accelerate a

Question

(5%) Problem 17: The world's fastest land animal, the cheetah, can accelerate at 8.3 m/s2 to a top speed of ven cheetah observes a passing gazelle traveling at vg- 19.5 m/s and begins to chase it. 31 m/s. A > 25% Part (a) How long, in seconds, does it take the cheetah to reach its maximum velocity, v max assuming its acceleration is constant? Grade Summary Deductions Potential 0% 100% cos) asin)acos0 Submissions Attempts remaining: in tan 200 % per attempt) detailed view sinhO atan acotan) coshOt0 cotanh0 0 END Degrees Radians DEL CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. 25% Part (b) How far, in meters, has the cheetah traveled, dmax,Vel, when it reaches it maximum velocity? 25% Part (c) Assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, Vemax and vg for the time, tc, it takes the chectah to catch the gazelle 25% Part (d) What is the numeric value for this time, Ic n seconds, assuming the cheetah is 28 m away when it reaches maximum velocity?

Explanation / Answer

Part A.

Given that

Max speed of Cheetah = 31 m/sec

acceleration of Cheetah = 8.3 m/sec^2

Vi = initial velocity = 0 m/sec

Now using equation

Vmax = Vi + a*t

t = (Vmax - Vi)/a

t = (31 - 0)/8.3

t = 3.735 sec

Part B.

Distance covered by cheetah during this time will be

d = Vi*t + 0.5*a*t^2

d = 0*3.735 + 0.5*8.3*3.735^2

d = 57.893 m = 57.9 m

Part C.

Now given that in time tc cheetah catches gazelle, then distance traveled by cheetah in this time would be equal to d + distance traveled by gazelle

distance traveled by cheetah = Speed of cheetah*time taken taken to catch

distnace travelled by gazelle = Speed of gazelle*time taken by gazelle

So,

vcmax*tc = d + vg*tc

Part D.

when d = 28 m

Using known values

31*tc = 28 + 19.5*tc

tc = 28/(31 - 19.5) = 2.43 sec

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