Vectors, how to find average acceleration

You should also find the direction.In summary, a cyclist is initially 20.5 m south of a flagpole in a park, moving east with a speed of 14.1 m/s. 3.95 seconds later, the cyclist is 36.3 m west of the flagpole, moving north with a speed of 18.3 m/s. Using this information, we can find the displacement, average velocity, and average acceleration of the cyclist in this interval. The displacement is approximately North-West, the average velocity is 10.6 m/s, and the average acceleration is 5.85 m/s/s.
  • #1
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..
 
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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
the answer i wrote was 3.23 m/s/s

ah! i will keep that in mind :)
 
  • #6
I think your vector subtraction must have gone wrong, as I do not get 3.23. Show your calculations.
 
  • #7
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
 
  • #8
juju1 said:
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.
 
  • #9
so - initial velocity for ihat = 14.1 and for jhat it = 18.3
what is final velocity?
 
  • #10
juju1 said:
initial velocity for ihat = 14.1
Yes.
juju1 said:
and for jhat it = 18.3
No, read the question again:
juju1 said:
going due east with a speed of 14.1 m/s.

juju1 said:
what is final velocity?
juju1 said:
3.95 s later... going due north with a speed of 18.3 m/s.
 
  • #11
so initial = 14.1 and final = 18.3
so change in velocity is inital - final over time
 
  • #12
juju1 said:
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.
 
  • #13
so its 5.85!
 
  • #14
juju1 said:
so its 5.85!
That is the magnitude, yes.
 

Related to Vectors, how to find average acceleration

1. What is a vector?

A vector is a mathematical object that has both magnitude (size) and direction. It is represented by an arrow pointing in a specific direction with a specific length.

2. How do you find the average acceleration of a moving object?

To find the average acceleration, you need to know the initial velocity, final velocity, and the time it took the object to change its velocity. The formula for average acceleration is (final velocity - initial velocity)/time.

3. What is the difference between average acceleration and instantaneous acceleration?

Average acceleration measures the overall change in velocity over a certain period of time, while instantaneous acceleration measures the change in velocity at a specific moment in time.

4. Can vectors be added or subtracted?

Yes, vectors can be added or subtracted using mathematical operations. When adding or subtracting vectors, their magnitudes and directions are taken into account.

5. How do you represent a vector mathematically?

A vector can be represented mathematically using coordinates or components. For example, a vector with a magnitude of 5 units pointing in the positive x-direction can be represented as (5, 0) or 5\hat{i}.

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