Finding when speed equals 10 m/s for particle with velocity v=(5.9t-4.1t²)i+8.7j

[tjbateh]

Homework Statement

The velocity v of a particle moving in the xy plane is given by v= (5.9 t - 4.1 t²)i + 8.7j, with v in meters per second and t (> 0) in seconds. (a) What is the acceleration when t = 3.7 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s?

Homework Equations

The Attempt at a Solution

Part A is -24.44 i + 0 j m/s²

Part B is .7195 seconds.

How do I find part C?

 

[rock.freak667]

if velocity is v=xi+yj

then |v| =speed = √(x²+y²)

 

[tjbateh]

That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?

 

[rock.freak667]

That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?

the 'y' term is a constant, so you can square both sides of the equation and then make the 'x²' the subject and then take the square root of both sides, then solve. (I hope you understood what I meant)

 

[tjbateh]

No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??

 

[rock.freak667]

No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??

ok we'd get


$$10=\sqrt{(5.9t-4.1t^2)^2 +(8.7)^2}$$


so if you square both sides you get rid of the square root sign. Then rearrange and make (5.9t-4.1t^2)² the subject and take the square root of both sides now.

 

[tjbateh]

would I get sometime like 1.3= t(5.9-4.1t)??