Solving for Rate of Change of Speed: 128° & 6.68 m/s2

[shiv604]

Homework Statement

An object is traveling along a curved path. At a particular instant the angle between its velocity vector and its acceleration vector is 128 degrees. The magnitude of its acceleration vector is 6.68 m/s2.

At that instant, what is the particle's rate of change of speed? (in m/s2)

Pick one:

a. 6.68
b. -6.68
c. -4.11
d. 4.11
e. -5.26
f. 5.26

Homework Equations

I am not sure which equation to use. That is where I am stuck. No equation I can think of helps me.

The Attempt at a Solution

Created a triangle, one side is the acceleration, the other is velocity. They are bound by a 128° angle. Not sure how to carry on from this point; help needed.

 

[Simon Bridge]

Don't worry about the equation - sketch the diagram.
Draw the two vectors.
Label the angle.
Continue the velocity with a dotted line.
resolve the acceleration against the velocity.

It's just trigonometry - but you have a bit of physics to do first.

The perpendicular component of the acceleration changes the ... what?
The parallel component changes the ...?

 

[shiv604]

Hmm. Well I made a triangle, one side velocity, the other acceleration; which was bound by 128 degrees.
I'm stuck on what to do next with this triangle.
I have never seen a problem like this before. I have tried my class notes, my textbook, and google. I cannot seem to grasp what the question is getting at. :(

 

[Simon Bridge]

How to describe it ... put the x-axis along the velocity vector and get the x and y components of the acceleration.

The component along the y-axis changes the direction of the velocity.
(Like in "circular motion".)

The component along the x-axis changes the magnitude, the speed.
(Like in "kinematics".)

 

[shiv604]

ok that makes sense!
I got 4.11 for x, and 5.26 for y.
I am confused on which one affects the rate of change.
Also once I do find which component affects the rate of change, would it be positive or negative?

 

[shiv604]

ooooh 4.11

 

[shiv604]

thanks a bunch!

 

[Simon Bridge]

Don't forget the units.

You now have two acceleration vectors - draw the arrows on your diagram.

You just need to realize that acceleration points in the same direction as force.
So the direction of the acceleration tells you the direction of the force.
You should be able to figure out which is which from that.

If the force would slow it down, then the change in velocity is negative.
If the force would speed it up, then the change in velocity is positive.

 

[shiv604]

yes thank you simon!

 

[Simon Bridge]

No worries: now you have a fancy new tool :)