Question

Determine the parametric equations of the position of a particle with constant velocity that follows a straight line path on the plane if it starts at the point P (9,7) and after one second it is at the point Q(−7,10) .

I need to find x(t), y(t) and the speed of the particle.

I know that vector PQ = <-16,3>

Would my parametric eqs be
x(t) = 9 -16t, and
y(t) = 7 + 3t ???

Answer 1

P=<9,7>
Q=<-7,10>

Answer 2

Q-P

Answer 3

Q-P = <-16,3> right?

Answer 4

Please someone provide a clear explanation

Answer 5

Yes you are right

Answer 6

Now p is your starting point,right

Answer 7

yes

Answer 8

<-16,3> is the direction

Answer 9

so my equations posted above are correct?

Answer 10

wouldn't it be 7 +3t?

Answer 11

So we add origin to direction
9-16t
7+3t

Answer 12

Can anyone explain this to me verbally with detail?

Answer 13

So if you plug in 1 for t you would get Q

Answer 14

So your are starting at P and moving in the direction of Q

Answer 15

I understand that, I am asking if my equations posted in the original post are correct, and if so, how do I determine the speed. I know to use t=1, and plug that in to both of my equations, but after that I am confused. Once I get the values for x(t) and y(t) I am unsure of how to find the speed.

Answer 16

Your equation is correct. To Find speed you will have to take a derivative
x(t) = 9 -16t, and
y(t) = 7 + 3t ???

x[t]'=-16
y[t]'=3