Answer:
Explanation:
As far as the displacement goes, we have 2 displacement vectors. If we didn't have the angles to deal with, this would be a much simpler process, but then that wouldn't be any fun at all, would it? I'll deal with the average speed first, then the displacement, which is a vector addition problem.
The average speed is found by adding together the distances the student traveled and then dividing this sum by the total time he spent traveling. If we are told that the student runs at 4.5 m/s for 3.0 minutes, we can use this to find out the distance he ran during that time interval. However, the units are not the same. We will find the distance the student traveled by convering the time to seconds.
3.0 minutes = 180 seconds, and
4.1 minutes = 246 seconds.
That means that the distance he ran in 180 seconds is found by multiplying this time be the speed at which he ran:
4.5 m/s(180 s) = 810 m and
3.5 m/s(246 s) = 860 m (rounded to follow the rules of sig dig).
This makes the speed equation look like this:
[tex]s=\frac{810+861}{180+246}=\frac{1671}{426}=3.9\frac{m}{s}[/tex] That's the average speed, which is NOT at all the same as the displacement. Displacement is where he ended up in reference to where he started. The angles play a huge part in this math (that is very involved, to say the least). We begin by restating the displacement of each "leg" of this journey.
The first leg took him 810 m at 207 degrees and
the second leg took him 860 m at 325 degrees
To find the x and y components of these 2 legs, or parts, we have to use the cos and sin formulas. We will call the first leg A and the second leg B. First the x components of both A and B:
[tex]A_x=810cos207[/tex] and
[tex]A_x=-720[/tex]
[tex]B_x=860cos325[/tex] and
[tex]B_x=704[/tex] and we add these to get the x-component of the resultant vector, C:
-720
+ 704
-10 (rounded, as needed, to the tens place).
Now for the y-components of the resultant vector:
[tex]A_y=810sin207[/tex] and
[tex]A_y=-370[/tex]
[tex]B_y=860sin325[/tex] and
[tex]B_y=-490[/tex] and we add these to get the y-component of the resultant vector, C:
-370
+ -490
-860
Since the x component is negative and so is the y, we are in QIII, so when we finally find our angle, we will have to add 180 to it.
For the magnitude of the displacement vector, in m:
[tex]C_{mag}=\sqrt{(-10)^2+(-860)^2}[/tex] which gives us
[tex]C_{mag}=860m[/tex]
Now, because displacement is vector, we also need the angle. We find that is the formula
[tex]\theta=tan^{-1}(\frac{C_y}{C_x})[/tex] and filling in:
[tex]\theta=tan^{-1}(\frac{-860}{-10})=90[/tex] (rounded correctly), and then we add 180 to give us a final direction of 270 degrees.
So the final displacement of the student is 860 m at 270 degrees
