Calculating Average Speed: Solving the v=8sin2t+3cos2t Equation

[Nicola Sterritt]

Hello,
I was wondering if someone out there could help me. The question:

The speed, v metres per second, of a particle is given as v=8sin2t+3cos2t
where t is in seconds.
i) Find the total distance traveled in the time interval 0≤t≤1
ii) Find the average speed over this time interval.

 

[BvU]

Hello Nicola,

This looks an awul lot like homework, and for that PF has a dedicated forum. With some rules: you need to show an attempt at solution and to use the template:

Homework Statement

The speed, v metres per second, of a particle is given as v=8sin2t+3cos2t
where t is in seconds.
i) Find the total distance traveled in the time interval 0≤t≤1
ii) Find the average speed over this time interval.​

Homework Equations

...
​

The Attempt at a Solution

...
​

Your turn! Help is on the way

 

[Nicola Sterritt]

Hello Nicola,

This looks an awul lot like homework, and for that PF has a dedicated forum. With some rules: you need to show an attempt at solution and to use the template:

Homework Statement

The speed, v metres per second, of a particle is given as v=8sin2t+3cos2t
where t is in seconds.
i) Find the total distance traveled in the time interval 0≤t≤1
ii) Find the average speed over this time interval.​

Homework Equations

...
​

The Attempt at a Solution

...
​

Your turn! Help is on the way

Here are the solutions:

 

[Nicola Sterritt]

Hi,

Believe it or not I am a teacher ! How embarrassing haha!
We are having a debate about the solution to this question!

I do not understand why the solution divides by 2 if you use the average value formula in integration to find average speed. If you substitute in values between 0 and 1 then the speed never goes above 3 point something so average speed can not be 7.03 but why divide by 2?

 

[BvU]

Speed never goes below 3. Note that $$ 8\sin(2t)+3\cos(2t) = \sqrt{73} \; \sin (2t+\phi) \quad {\rm \ with \ \ } \cos\phi = {8\over \sqrt{73}} $$

The division by 2 is a mistake by the author of the book.

 

[Nicola Sterritt]

Hi,

Thank you. You may be able to help with one other question.

Find the general solution solution of the equation sin2x=-√3/2 and use it to find all the solutions for 0°≤x≤360°.

The solutions say that the answers are 120°,150°,300°,330°. I think that the solution set should be 0°≤2x≤360° for these to be the answers and that the answers should only be 120° and 150°.

Thank you for your help.
Nicola

 

[Samy_A]

Hi,

Thank you. You may be able to help with one other question.

Find the general solution solution of the equation sin2x=-√3/2 and use it to find all the solutions for 0°≤x≤360°.

The solutions say that the answers are 120°,150°,300°,330°. I think that the solution set should be 0°≤2x≤360° for these to be the answers and that the answers should only be 120° and 150°.

Thank you for your help.
Nicola

The solutions 120°,150°,300°,330° seem correct. These angles all lie between 0° and 360° and the sine of twice each of these angles is -√3/2.

 

[Nicola Sterritt]

Sorry, I got confused with another question I was doing. The question was
Find the general solution solution of the equation sin2x=-√3/2 and use it to find all the solutions for 0°≤x≤720°. ...NOT 360 as I had above.

The solutions say that 120°,150°,300° and 330° but I think that the solutions should include 480°, 510°, 660° and 690° also unless the solution set is changed to 0°≤2x≤720° or 0°≤x≤360°

 

[Samy_A]

Sorry, I got confused with another question I was doing. The question was
Find the general solution solution of the equation sin2x=-√3/2 and use it to find all the solutions for 0°≤x≤720°. ...NOT 360 as I had above.

The solutions say that 120°,150°,300° and 330° but I think that the solutions should include 480°, 510°, 660° and 690° also unless the solution set is changed to 0°≤2x≤720° or 0°≤x≤360°

I agree with you.