Physics, 03.07.2021 18:40 jonloya264

URGENT A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes. a. What was his average speed? b. What was his displacement? PLEASE SHOW ALL WORK

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Answer from: ereyes181

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:

$s=\frac{810+861}{180+246}=\frac{1671}{426}=3.9\frac{m}{s}$ 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:

A_x=810cos207 and

A_x=-720

B_x=860cos325 and

B_x=704 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:

A_y=810sin207 and

A_y=-370

B_y=860sin325 and

B_y=-490 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:

$C_{mag}=\sqrt{(-10)^2+(-860)^2}$ which gives us

$C_{mag}=860m$

Now, because displacement is vector, we also need the angle. We find that is the formula

$\theta=tan^{-1}(\frac{C_y}{C_x})$ and filling in:

$\theta=tan^{-1}(\frac{-860}{-10})=90$ (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

Answer from: Quest

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction

Answer from: Quest

That big glowy thing in the sky is called the sun

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URGENT A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s...