Can someone help me with this algebraic problem??

Question

"Playing with her dog in a river, a woman threw a rubber ball downstream. Starting immediately from where the woman stood, the dog caught up to the ball in a minute and a half, but took two and a half minutes for the dog to swim back with the ball. How long would it have taken for the ball to float from the woman to the point where the dog caught up to it.?"

I will be giving the points to the person with the most helpful answer that is explained the best to me. Thanks so much for your help!!

Answer 1

Variables:

d = distance the ball is thrown (feet)

r = rate of flow of the river (feet per minute)

p = rate of dog swimming in still water (feet per minute)



When the ball floats downstream, it rate will be p, the same as the river.

When the dog swims downstream in the river, the rate of the river will be added to his ability to swim in still water, so his rate downstream will be p+r.

When he swims upstream, the rate of the river will be subtracted from his ability to swim in still water, so his rate upstream will be p-r.



The key formula is (distance) = (rate)*(time).



The distance the dog swims downstream is

1.5*(p+r)

The distance the dog swims upstream is

2.5*(p-r)

The dog swims the same distance both ways, so

1.5*(p+r) = 2.5*(p-r)

1.5*p + 1.5*r = 2.5*p - 2.5*r

2.5r + 1.5*r = 2.5*p - 1.5*p

4*r = 1*p



Therefore (p+r) = 4r + r = 5r.

This means the dog can swim downstream 5 times as fast as the ball floats. So the ball would take 5 times as long to go the same distance.



The dog took 1.5 minutes to swim downstream to the point where he catches the ball.

So it would have taken the ball 5*1.5, or 7.5 minutes, to float that same distance.



It's interesting that the distance d that the ball was thrown was never needed in this solution. If we wanted it, it's clear that the ball floated only 1/5 of the distance it traveled. So it must have traveled 4/5 of the distance in the air.

Answer 2

Let Vs = velocity of stream is units/minute



Let Vd = velocity of dog swimming in STILL WATER



It is assume that the velocity of the stream and dog are additive



That is:



Velocity of dog relative to woman is Vd+Vs when swimming downstream

Velocity of dog relative to woman is Vd-Vs when swimming upstream



Now let D be the distance from the women when the ball hits the water downstream.



The ball is assumed to float downstream at Vs.



Let Dc be the distance from the woman where the dog catches the ball.



Thus wee want to find time T = Dc/Vs which is the time it will

take the ball to float from the woman to the distance it was caught at.





So, when the dog is swimming downstream, it is chasing the ball

also floating downstream. The velocity of the dog relative to the

ball is Vd+Vs-Vs = Vd.



Thus if it takes the dog 1.5 minutes to cover the distance D:



Vd = D / 1.5 (1)



Similarly, it takes dog 2.5 minutes to return:



Vd-Vs = Dc / 2.5 (2)



As well, we know that during the 1.5 minutes to catch the

ball, the ball moved an additional 1.5 * Vs units further downstream:



Dc = 1.5Vs + D (3)



We have 3 equations, and 4 unknowns (Vd, Vs, D, Dc). We need to eliminate

D and Vd



From (2)



Vd = Dc/2.5 + Vs



Subing into (1) and re-arranging:



D = 1.5(Dc/2.5 + Vs)



Subing into (3) gives



Dc = 1.5Vs + 1.5(Dc/2.5 + Vs)



from which we get:



0.4Dc = 3Vs



From which we get:





Dc/Vs = 7.5 minutes

Answer 3

Let d = distance from woman to point where dog catches ball,

v_w = velocity of surface of water, and v_d = velocity of dog on still water. Then:



v_d + v_w = d/(3/2) = 2d/3, [1] (dog swims downstream to ball)

v_d - v_w = d/(5/2) = 2d/5, [2] (dog swims upstream back to woman)

[2] x (-1): - v_d + v_w = - 2d/5, [3]

[1] + [3]: 2v_w = 2d/3 - 2d/5 = 4d/15,

v_w = 2d/15.



Time taken for ball to float from woman to catching point is time taken for surface of water to move from woman to catching point:



d/v_w = d/(2d/15) = 15/2 = 7.5 min.

Answer 4

How strong is the friction of the current?