# Average speed question

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1. Jan 26, 2016

### 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 travelled in the time interval 0≤t≤1
ii) Find the average speed over this time interval.

2. Jan 26, 2016

### 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:

1. The problem statement, all variables and given/known data
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 travelled in the time interval 0≤t≤1
ii) Find the average speed over this time interval.​

2. Relevant equations
...
3. The attempt at a solution
...
Your turn! Help is on the way

3. Jan 26, 2016

### Nicola Sterritt

Here are the solutions:

4. Jan 26, 2016

### 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?

5. Jan 26, 2016

### 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.

Last edited: Jan 26, 2016
6. Jan 26, 2016

### 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

7. Jan 26, 2016

### Samy_A

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.

8. Jan 26, 2016

### 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°

9. Jan 26, 2016

### Samy_A

I agree with you.