Answer:
a) t=0.136hrs, b) t=0.3 hrs, c) ∠=112.02°, d) t= 0.108hrs, e) t=0.1hrs and x=0.3km
Explanation:
a)
In order to solve each of the parts of this problem, we must start by drawing a diagram of each situation. (See picture attached.) That way it will be easier for us to analyze the problem.
Once we drew the situation we can start modeling it. As you may see, both velocities are going in the same direction, so they will be added to find the total velocity of the boat:
[tex]V_{t}=V_{b}+V_{r}[/tex]
where:
[tex]V_{t}[/tex] is the total velocity of the boat.
[tex]V_{b}[/tex] is the the velcity of the boat relative to the water and
[tex]V_{r}[/tex] is the velocity of the river.
So we can find the total velocity:
[tex]V_{t}=8.0km/hr+3.0km/hr[/tex]
[tex]V_{t}=11.0km/hr[/tex]
once e got this value, we can use the following equation:
[tex]v=\frac{x}{t}[/tex]
and solve it for t:
[tex]t=\frac{x}{v}[/tex]
So we can substitute the values we know:
[tex]t=\frac{1.5km}{11km/hr}[/tex]
which yields:
t=0.136hrs
(this answer can be turned to minues and seconds if specified by the problem)
b)
Part b is similar, with the difference that the boat is rowing in the opposite direction of the river, so the answer is slightly different:
Both velocities are going in opposite directions, so they will be subtracted to find the total velocity of the boat:
[tex]V_{t}=-V_{b}+V_{r}[/tex]
So we can find the total velocity:
[tex]V_{t}=-8.0km/hr+3.0km/hr[/tex]
[tex]V_{t}=-5.0km/hr[/tex]
once e got this value, we can use the following equation:
[tex]v=\frac{x}{t}[/tex]
and solve it for t:
[tex]t=\frac{x}{v}[/tex]
So we can substitute the values we know:
[tex]t=\frac{-1.5km}{-5.0km/hr}[/tex]
which yields:
t=0.3hrs
c)
On part c we can see that the sum of the velocities should give us a velocity perpendicular to the velocity of the river. This will allow us to use trigonometric functions and ratios to solve for the desired angle.
As you may see, a triangle is formed with the vectors of velocity, so we can use them to find the desired angle. We know that:
[tex]cos (\theta)=\frac{adjascent}{hypotenuse}[/tex]
in this case my adjascent velocity is 3km/hr and the hypotenuse is 8km/hr, so we can substitute them into the given equation:
[tex]cos (\theta)=\frac{3km/hr}{8km/hr}[/tex]
and solve for theta:
[tex] \theta=cos^{-1}(\frac{adjascent}{hypotenuse})[/tex]
which yields:
[tex]\theta=112.02^{o}[/tex]
d)
On part d, they want us to find the time it takes for the boat to go through the 0.8km river, and we can find that by using pythagore.
First, we find the total velocity:
[tex]V_{b}^{2}=V_{r}^{2}+V_{t}^{2}[/tex]
So now we can solve that for the total velocity:
[tex]V_{t}=\sqrt{V_{b}^{2}-V_{r}^{2}}[/tex]
and we can substitute the given data:
[tex]V_{t}=\sqrt{(8km/hr)^{2}-(3km/hr)^{2}}[/tex]
Which yields:
[tex]V_{t}=7.42km/hr[/tex]
With this magnitude, I can proceede and find the desired time:
[tex]t=\frac{y}{v}[/tex]
so the time will be:
[tex]t=\frac{0.8km}{7.42km/hr}[/tex]
when solving we get:
t=0.108hrs.
e)
On part e, the bot will be moving both to the right and to the upper part of the diagram. In this case we can start by finding the time it takes the boat to reach the other side of the river. We can do so like this:
[tex]t=\frac{y}{v}[/tex]
so the time will be:
[tex]t=\frac{0.8km}{8km/hr}[/tex]
when solving we get:
t=0.1hrs.
In order to fing how far away the boat will go to the right, we can use the following formula:
[tex]x=v_{r}t[/tex]
so we get:
x=(3km/hr)(0.1hr)
x=0.3km
