Question

The position vector of a particle moving in the xy-plane at time t is given by p= (3t^2-4t)i+ (t^2+2t)j. The speed of the particle at t=2 is

Answer 1

speed = || s'(t) ||

Answer 2

i got the derivative of (6t-4)i+(3t^2-4t)+(2t=2)j+(t^2+2t)
what do i do with the o and j?

Answer 3

nothing

Answer 4

but my answer choices does;t involve any variable.

Answer 5

plug t = 2 in and you'll get a number

Answer 6

Is 10 one of your answers?

Answer 7

@eliassaab yes!! but how??

Answer 8

\[
p(t) =(3 t^2 - 4 t, t^2 + 2 t)\\
p'(t)=( 6 t-4, 2t +2)\\
p'(2)={8,6}\\
||p'(2)||= \sqrt{ 8^2 +6^2}=\sqrt {100}=10
\]

Answer 9

p'(2)=(8,6) = 8 i + 6j

Answer 10

Where are you?

Answer 11

it is asking the speed, so won't i just find the velocity and plug in??
you used the distance formula, right?

Answer 12

The speed is the length of the velocity vector. The length of a i + b j is
\[\sqrt{a^2 + b^2}
\]

Answer 13

oh, thanks!!

Answer 14

YW