An alligator is stalking its zebra prey located 20 meters upstream of a river, o

Question

An alligator is stalking its zebra prey located 20 meters upstream of a river, on the opposite side of the river. the time T it takes for the alligator to reach its prey can be minimized if the alligator swims to a point P, x meters upstream, on the other side of the river.

The river is 6 meters wide, and the water's velocity is negligible. The speed vL of the alligator on land is 0.25m/s and vW in the water is 0.20m/s.

1. Calculate the time taken T1if the alligator does not travel on land.

2. Calculate the time taken T2 if the alligator swims the shortest distance possible.

3. Write the general equation of the total time T that it takes the alligator to reach the zebra in terms of the distance x.

4. Between these two extremes, there is one value of x which minimizes the time taken. Calculate this value of x, and, hence, calculate the minimum possible time for the alligator to enjoy a zebra meal!

Explanation / Answer

1.

In this case alligator travels distance ‘r’

r= sqrt(6^2 +20^2) 20.88 m

T1=r/0.20

T1= 20.88/0.25

T1=82.52s

2.

In this shortest distance = 6.0m in the river.

And the distance on the land = 20.0m

T2=6/0.20+20/0.25

T2=110s

3.

T= time in water + time on land

T= [sqrt(6^2+x^2)]/0.20+ (20-x)/0.25 -----------(1)

4.

T= [sqrt(6^2+x^2)]/4.0+ (20-x)/5.0

T=5*(6^2+x^2)^1/2 + 80-4x

Take derivative of above eqn,

T ‘= 5x/sqrt(36+x^2) – 4

To minimize x,

T’=0

Hence,

0 = 5x/sqrt(36+x^2) – 4

Gives,

x= 8.0 m

Plug this values in eqn(1),

T=[sqrt(6^2+8^2)]/0.20+ (20-8)/0.20

T = 98.0s

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