# Math Help - Velocity and Position Problem

1. ## Velocity and Position Problem

A particle moves along the x-axis so that at any time t>0, its velocity is given by $v(t)=4-6t^2$. If the particle is at position x=7 at time t=1, what is the position of the particle at time t=2?
(a) -10
(b) -5
(c) -3
(d) 3
(e)17

How would I figure out the position of the particle when I'm only given the velocity equation and not the position equation? I tried integrating the velocity equation to get the position equation, but when I plug back in t=1, the position equation that I get from integrating doesn't give me x=7.

2. You have to remember your constant of integration (that + c). The question gives you the one point (1, 7) so that you can calculate that constant.

4t - 2t^3 + c

4(1) - 2(1)^3 + c = 7
c = 5

4t - 2t^3 + 5

3. Originally Posted by summermagic
A particle moves along the x-axis so that at any time t>0, its velocity is given by $v(t)=4-6t^2$. If the particle is at position x=7 at time t=1, what is the position of the particle at time t=2?
(a) -10
(b) -5
(c) -3
(d) 3
(e)17

How would I figure out the position of the particle when I'm only given the velocity equation and not the position equation? I tried integrating the velocity equation to get the position equation, but when I plug back in t=1, the position equation that I get from integrating doesn't give me x=7.
$x(2) - x(1) = \int_1^2 v(t) \, dt$

so ...

$x(2) = 7 + \int_1^2 4 - 6t^2 \, dt$

4. Integrate v(t) to get the position s(t)

s(t) = 4t - 2*t^3 +c yes the int constant is imporant!

s(1) = 7 so 4-2+c = 7 so c= 5

s(t) = 4t - 2*t^3 + 5

now finish