- #1
pinkyjoshi65
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Two swimmers, Al and Bob, live on opposite shores of a 200.0 m wide river that flows east at 0.70 m/s. Al lives on the north shore and Bob lives on the south shore. They both set out to visit a mutual friend, Ted, who lives on the north shore at a point 100.0 m upstream from Al and 100.0 m downstream from Bob. Both swimmers can swim at 1.4 m/s through the water. How much time must Al wait after Bob sets out so that they both arrive at Ted's place at the same time? Both swimmers make their trips by the most direct routes.
So, first I can find the time taken by Al.
t=D/V_ag
= 100/V_aw-V_wg
=100/0.7= 142.85 Seconds
Now, for Bob
we could use trig to find the angle (beta) of his most direct route.
but 1st we need to find theta.
So using tri==== Tan(theta)= 200/100
And we get theta as 63.4 degrees.
Then we can find beta, since one angle is 90 and the other is 63.4.
so beta will be= 26.5 degrees
Then, Sin(26.5)= 0.7/hypotunese
hence hypotunese= 1.56m/s
This is bob's speed (most direct route)
Now for his time, t= 200/1.56= 128.20 seconds.
T1-T2= 142.82-128.20= 14.62 seconds.
hence Al must wait for 14.62 seconds after Bob has left inorder for both of them to reach Ted's place at the same time.
Does this look good?
So, first I can find the time taken by Al.
t=D/V_ag
= 100/V_aw-V_wg
=100/0.7= 142.85 Seconds
Now, for Bob
we could use trig to find the angle (beta) of his most direct route.
but 1st we need to find theta.
So using tri==== Tan(theta)= 200/100
And we get theta as 63.4 degrees.
Then we can find beta, since one angle is 90 and the other is 63.4.
so beta will be= 26.5 degrees
Then, Sin(26.5)= 0.7/hypotunese
hence hypotunese= 1.56m/s
This is bob's speed (most direct route)
Now for his time, t= 200/1.56= 128.20 seconds.
T1-T2= 142.82-128.20= 14.62 seconds.
hence Al must wait for 14.62 seconds after Bob has left inorder for both of them to reach Ted's place at the same time.
Does this look good?