# Dynamics with complex equations

1. Sep 16, 2010

### Ry122

Problem:
A particle starts moving from the position
r= -2i+j

with its velocity given as a function of time, t, as

v=(5t^2)i+2j
.
What is the distance of the particle from the origin at time t = 10 s?

The i's a j's both have hats if that makes any difference.

My attempt:
distance = velocity x time
so multiply the v equation by t and then subsitute 10s into it.
This will give the distance travelled in both the x and y direction add these values to the values given in the first equation.
So (5t^2)*t + t*2j

then sub in 10 and add the i and j values to their respective values in the first equation to find the final displacement.

I'm unsure if this is correct or not.

2. Sep 16, 2010

### gabbagabbahey

No, "displacement", not "distance" equals velocity x time, and even that is only true in the case of constant velocity.

Instead, you will want to integrate the velocity to get the position as a function of time, and then determine the distance from the origin at a given time by using the usual formula for the distance between 2 points.

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3. Sep 16, 2010

### Ry122

is this not exactly what you're telling me to do?
(5t^2)*t + t*2j

4. Sep 16, 2010

### gabbagabbahey

No, do you know how to integrate?

5. Sep 16, 2010

### Ry122

Sorry didn't read what you first said properly.
So once the integral is found, and the equation becomes something which represents the displacement of the object, can't I just sub 10 into this equation and then add -2i and 1j to the resulting i and j values?

6. Sep 17, 2010

### gabbagabbahey

Sort of, your integration (like any integration) will have a constant on the end. In this case, the constant will be a vector and equal to the initial position of the particle:

$$\int_0^t \textbf{v}(t')dt' = \int_0^t \frac{d\textbf{r}(t')}{dt'}dt' = \textbf{r}(t)-\textbf{r}(0)$$