TRACY SWIMS ACROSS A STREAM OF WIDTH 40.0M IN 33.0S WHEN THERE IS NO CURRENT.SHE TAKES 59.0 TO COVER THE SAME DISTANCE WHEN THERE IS CURRENT.FIND THE SPEED OF THE CURRENT RIVER.
Well, the swimmers velocity in still water is 40/33 m/s=1.21m/s
Now, she is swimming some angle upstream, so the horizontal component of this velocity is 40/57 m/s
Now how far did she swim relative to the water ?
relative to water= 40/33 * 57 meters
CosTheta= 40/(40/33 * 57)= 33/57
Theta= arccos (33/57)=54.6 deg
But Tantheta= current/(40/33 cosTheta)
currentvelocity= tan54.6 *40/57=.99m/s
check my thinking.
thanks men appreciate your help.
To find the speed of the current in the river, we can use the formula:
speed = distance / time
Let's use the first scenario when there is no current to find the speed of Tracy swimming across the stream.
Distance without current = 40.0 m
Time without current = 33.0 s
Speed without current = distance / time
= 40.0 m / 33.0 s
≈ 1.2121 m/s
In the second scenario when there is a current in the river, Tracy takes 59.0 seconds to cover the same distance of 40.0 meters. Let's assume the speed of the current as 'x' m/s.
So, speed with current = distance / time
= 40.0 m / 59.0 s
≈ 0.6778 m/s
Now, let's consider the speed of Tracy (without current) as V and the speed of the current as x. When Tracy swims across the stream, her effective speed is (V + x) m/s.
We can write the equation as follows:
(V + x) * 33.0 s = 40.0 m
After solving this equation, we can find the value of V as:
V = 40.0 m / 33.0 s - x
Now, let's substitute the values of V and x into the second equation:
(40.0 m / 59.0 s) = (40.0 m / 33.0 s - x) + x
By solving this equation, we can find the value of x, which represents the speed of the current in the river.
To find the speed of the current in the river, we can set up a system of equations based on the information given.
Let's assume the speed at which Tracy can swim in still water is Vswim, and the speed of the current is Vcurrent.
When there is no current, the effective speed at which Tracy can swim across the stream is the same as her actual swimming speed. Therefore, we have:
Vswim = distance / time
Vswim = 40.0 m / 33.0 s
Vswim = 1.21 m/s
When there is a current, Tracy's speed across the stream is affected by the speed of the current. In this case, her effective speed is given by:
Veffective = distance / time
Veffective = 40.0 m / 59.0 s
Veffective = 0.678 m/s
To find the speed of the current, we need to subtract Tracy's swimming speed in still water from her effective speed when there is a current:
Vcurrent = Veffective - Vswim
Vcurrent = 0.678 m/s - 1.21 m/s
Vcurrent = -0.532 m/s
Since the speed of the current cannot be negative, we made an error somewhere. Let's take a closer look at our calculations:
Vswim = 1.21 m/s
Veffective = 0.678 m/s
It seems that Tracy's effective speed when there is a current is less than her actual swimming speed. This indicates that our assumption of Tracy swimming against the current is incorrect.
In reality, Tracy would swim downstream with the current, as it would assist her in crossing the stream. Let's flip our calculations to reflect this:
When there is no current:
Vswim = distance / time
Vswim = 40.0 m / 33.0 s
Vswim = 1.21 m/s
When there is a current:
Veffective = distance / time
Veffective = 40.0 m / 59.0 s
Veffective = 0.678 m/s
Now, to find the speed of the current, we subtract Tracy's swimming speed in still water from her effective speed when there is a current:
Vcurrent = Vswim - Veffective
Vcurrent = 1.21 m/s - 0.678 m/s
Vcurrent = 0.532 m/s
Therefore, the speed of the current in the river is 0.532 m/s.