Question

Particle Postion!

The position of a particle as a function of time is given by r=( 6.60 i^+ 2.70 j^)t^2 m, where t is in seconds. What is the particle's distance from the origin at t = 3.00s ?

Answer 1

nah, just use pythagoreas and find the magnitude of the position vector r @ t = 3.

Answer 2

right, because its asking for the displacement .
if you wanted to find total distance traveled from t=0 to t = 3 that would be more involved. you would have to use the derivative, take magnitude of it.
you should do it for a challenge problem

Answer 3

Right

Answer 4

@nsakhamuri Is this equation what you wanted :
$$
\Large r(t)= (6.60 \hat{\imath}+ 2.70 \hat{\jmath}) t^2 \\
\Large r(t)= 6.60t^2 \hat{\imath}+ 2.70t^2 \hat{\jmath}


$$

Answer 5

@perl so its asking just for the particle's distance from the origin at t=3

Answer 6

correct

Answer 7

$$ \Large || r(3)|| $$

Answer 8

i got 83.7 but mastering physics says its wrong

Answer 9

$$
\Large r(t)= (6.60 \hat{\imath}+ 2.70 \hat{\jmath}) t^2 \\
\Large r(t)= 6.60t^2 \hat{\imath}+ 2.70t^2 \hat{\jmath}\\
\Large r(3)= 6.60\cdot 3^2 \hat{\imath}+ 2.70 \cdot 3^2 \hat{\jmath}\\
\Large r(3)= 59.4 \hat{\imath}+ 24.3 \hat{\jmath}\\
\Large ||r(3)||= \sqrt{59.4^2+ 24.3^2}\\


$$

Answer 10

Oh so you can't just add 6.6 and 2.7 and multiply it by 3^2?

Answer 11

no, they are independent vectors :)

Answer 12

And how would you find the speed with the same stuff?

Answer 13

the speed at t = 3 is


$$

\Large speed = ||r'(3)||
$$

Answer 14

how would you find the derivative for this type of function?

Answer 15

$$
\Large r(t)= 6.60t^2 \hat{\imath}+ 2.70t^2 \hat{\jmath}\\
\Large r'(t)= 6.60 \cdot 2t~ \hat{\imath}+ 2.70 \cdot 2t~ \hat{\jmath}\\
\Large r'(t)= 13.2~t \hat{\imath}+ 5.4~t \hat{\jmath}\\
\Large r'(3)= 13.2(3) \hat{\imath}+ 5.4(3) \hat{\jmath}\\
\Large r'(3)= 39.6 \hat{\imath}+ 16.2 \hat{\jmath}\\
\Large ||r'(3)||= \sqrt{39.6^2+ 16.2^2}\\


$$

Answer 16

ohh okay thanks soo much @perl