## anonymous 4 years ago The position of a particle as a function of time is given by r_vec =( 6.60 \hat{ i }+ 2.40 \hat{ j })t^{2}\;{\rm m}, where t is in seconds. What is the particle's speed at t_1 = 2.20 s?

1. anonymous

Take the derivative with respect to time of r-vector to get the velocity-vector. To take the derivative, take the derivative of each dimensional component. For instance, the x component 6.60's derivative is 0, so the velocity's x component is 0. The y component 2.4t^2 can be differentiated to to 4.8t. Being that it's a vector, you should create a right triangle with both components and solve for the hypotenuse (final vector), but there's no x component so just plug in 2.2 for t in 4.8(t) for your answer.

2. anonymous

someone else posted this question or similar. anyway, basically didnt read the previous answer but i did see derivative. which is what u need to do. first derivative. then just plug in ur t and getv :) eazy peazy :)

3. anonymous

that didnt work. i had tried that already. when you take the derivative of (6.6+2.4)t^2 you get 13.2t+4.8t which would be your velocity at at 2.2 seconds your speed should be 39.6. BUT doing this on mastering phyusics it tell me the answer is incorrect.

4. anonymous

You're multiplying wrong. I got a solution of 10.56.

5. anonymous

Actually, that was an unhelpful answer. Give me some time to do it all out.

6. anonymous

7. anonymous

which equates to $\sqrt{6.6^2 + 2.4^2} *2.2 * 2$ where 2.2= t and 2 came from the derivative

8. anonymous

9. anonymous

10. anonymous

This should be able to solve any "find the velocity given this position vector" in 2D.

11. anonymous

$r=(6.6 i + 2.4 j)t^{2}$