Question

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?

Answer 1

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.

Answer 2

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 :)

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

the correct answer is 30.9

Answer 7

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

Answer 8

Huh, could you type out your problem without using LaTeX? Maybe it copied badly in your asking of it.

Answer 9

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

Answer 10

\[r=(6.6 i + 2.4 j)t^{2}\]