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Simple Vector Alegebra Problem

  • #1
222
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Hello all. I have a problem from my homework that I can't seem to figure out. It is a problem from a freshman introductory physics course. Here goes:
The position of a particle as a function of time given by: r =(5i^+4j^)t^2m , where t is in seconds. Find an expression for the particles velocity v as a function of time.

Where I wrote the ^ after the letters means its a unit vector.
Now isn't the way that one would find the velocity is to take the derivative of the expression? Or is that just for instantanious velocity? And if the way to find to velocity is to take the derivative, do I first distribute the t^2 then differentiate, ending up with: f'(t)=(10ti^+8tj^)m/s? Or do I use the product rule and end up with: f'(t)=(9t^2+10ti^+8tj^)m/s?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
1
You take the derivatve. It doesn't matter if you distribute first, the coefficient of [itex]t^2[/itex] is a constant.
 
  • #3
222
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So the answer is v = f'(t) = (10ti^+8tj^)m/s ?
 
  • #4
LeonhardEuler
Gold Member
859
1
yup.
additional charaters to make the post 10 characters long
 
  • #5
222
0
LeonhardEuler said:
additional charaters to make the post 10 characters long
I don't understand what you mean by this.
 
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