Vectors, how to find average acceleration

[juju1]

Homework Statement

At one instant a bicyclist is 20.5 m due south of a park's flagpole, going due east with a speed of 14.1 m/s. Then, 3.95 s later, the cyclist is 36.3 m due west of the flagpole, going due north with a speed of 18.3 m/s. For the cyclist in this 3.95 second interval, find each of the following.

Homework Equations

(a) displacement magnitude and direction:
(b) average velocity magnitude and direction:
(c) average acceleration magnitude and direction:

The Attempt at a Solution

a)So i found the displacement by the formula sqrt(20.5ihat^2 + 36.3jhat^2) but for some reason i couldn't find the direction. i drew it out, and my resultant vector points NorthWest, i also plugged in inversetan(36.3/20.5) and it was wrong
b) I found the average velocity by taking 20.5ihat and 36.3jhat and dividing that by the time and then plugging it into the equation and got (10.6m/s)...once again, i don't know how to find the direction
c) I am VERY confused on how to find the av.acceleration, i tried taking the average velocity that i found by dividing 20.5ihat and 36.3jhat and dividing that by time but it was wrong..

 

[andrewkirk]

For (c), average acceleration is change in velocity divided by time.
So, to find the average acceleration, just subtract the initial velocity vector from the final velocity vector to get a 'change of velocity' vector. Then divide that by the time.

For (b), just divide your answer to (a) by the time, as average velocity is displacement (change in location) divided by time.

For (a), your magnitude is correct. The displacement is approximately North-West, but not exactly. It is certainly related to $\tan^{-1}\frac{36.3}{20.5}$ but it depends on how you wrote your answer. That inverse tan gives you the degrees to the West of North.

 

[juju1]

so for C, i tried doing what you said, but it was wrong. my final velocity would be what i found the get the average velocity right? (that ihat and jhat) and my initial would be what they give me in the problem?
also it doesn't state what direction i need it into be..but inverse(36.3/20.5) is wrong

 

[andrewkirk]

Show exactly what you wrote for the answer.

By the way, it's best to say 'it didn't match the assigned answer' rather than 'it is wrong'. Assigned answers are themselves incorrect on occasion.

 

[juju1]

the answer i wrote was 3.23 m/s/s

ah! i will keep that in mind :)

 

[andrewkirk]

I think your vector subtraction must have gone wrong, as I do not get 3.23. Show your calculations.

 

[juju1]

average velocity=10.6m/s
we got that by dividing position by time
so i took the numbers we got from position divided by time (5.189ihat and 9.189jhat) and divided those AGAIN by the time and got 1.759 and 3.115. did the equation sqrt(1.759^2 and 3.115^2) to get the answer 3.23 m/s/s

 

[andrewkirk]

average velocity=10.6m/s
we got that by dividing position by time

Average velocity plays no part in the calculation of average acceleration.
Review the first line of post 2 above.

 

[juju1]

so - initial velocity for ihat = 14.1 and for jhat it = 18.3
what is final velocity?

 

[haruspex]

initial velocity for ihat = 14.1

Yes.

and for jhat it = 18.3

No, read the question again:

going due east with a speed of 14.1 m/s.

what is final velocity?

3.95 s later... going due north with a speed of 18.3 m/s.

 

[juju1]

so initial = 14.1 and final = 18.3
so change in velocity is inital - final over time

 

[haruspex]

so initial = 14.1 and final = 18.3
so change in velocity is inital - final over time

Yes, but these are velocities, so they have direction, and the subtraction is to be done vectorially.
It is exactly the same method you used to find average velocity, but using an initial and final velocity instead of an initial and final position.

 

[juju1]

so its 5.85!

 

[haruspex]

so its 5.85!

That is the magnitude, yes.