# Find time at which position vector points towards origin

1. Jul 26, 2016

### Taniaz

1. The problem statement, all variables and given/known data
The position vector of a particle at time t is R=(1-t^2)i+(3t-5t^2)j. Find the time at which P is moving (a) towards the origin (b) away from the origin.

2. Relevant equations

3. The attempt at a solution

I've thought about this for a while but I've come to the conclusion that I'm not really sure what they're asking for when they say towards and away from the origin. I know that the acceleration is constant because if you keep deriving you get a = -2i - 10j

2. Jul 26, 2016

### RUber

They are asking for a time when the velocity $R'$ is such that given $R = x \hat i + y \hat j$, $R'$ is pointing in the direction of the origin.
For example, if R = (x,y), what would a velocity vector pointing toward the origin look like?

3. Jul 26, 2016

### Taniaz

How did you deduce that? The velocity vector would be tangent to R?

4. Jul 26, 2016

### haruspex

The velocity vector is the time derivative of the position vector by definition. Similarly, acceleration is the derivative of velocity.

5. Jul 26, 2016

### Taniaz

Yes I do know that but I'm not sure what it looks like pointing towards and away from the origin.

6. Jul 26, 2016

### haruspex

In general, when we discuss vectors, they only have magnitude and direction. They do not have a specific anchor point. (Forces do have a point of application as well as being vectors.) But in this question, you need to think of the velocity as being a vector from the current position.

7. Jul 27, 2016

### Taniaz

The position vector R starts from the origin so supposing it lies at some point (x,y) at time t, pointing away from the origin, so will we draw it from the tip of this position vector? If it's towards the origin, will it be pointing downwards from the tip?

8. Jul 27, 2016

### haruspex

Yes, except that I'm not sure what you mean by 'downwards' there. It will be pointing from current position towards origin.

9. Jul 27, 2016

### Taniaz

If it is pointing from the current position to the origin, then it will just be coinciding with the position vector in the opposite direction?

10. Jul 27, 2016

### haruspex

Yes.

11. Jul 27, 2016

### Taniaz

But how does this tell us about time? We find the derivative of x to get velocity and equate it to negative of the position vector?

12. Jul 27, 2016

### haruspex

Yes, but only the direction, not the magnitude

13. Jul 27, 2016

### Taniaz

But that doesn't make sense because if you equate v= -R, how do you solve for t?

14. Jul 27, 2016

### haruspex

You know the position at time t and the velocity at time t.

15. Jul 27, 2016

### Taniaz

Do we have to find an expression or a definite value for t?

16. Jul 27, 2016

### haruspex

Two definite values.

17. Jul 27, 2016

### Taniaz

We do get 2 values for t for both the i and j components because it's a quadratic equation but for the one with towards the origin, the solutions aren't whole numbers I think

18. Jul 27, 2016

### haruspex

19. Jul 27, 2016

### Taniaz

v=-R and v=(-2t)i + (3-10t)j
(-2t)i + (3-10t)j = -(1-t^2)i - (3t-5t^2)j
(-2t)i + (3-10t)j = (-1+t^2)+(-3t+5t^2)j

(-2t)i = (-1+t^2)i and (3-10t)j=(-3t+5t^2)j
t = -1-sqrt2 or t = sqrt 2-1 for the i components and
t= (-7- sqrt 109)/10 or t =sqrt 109-7/10

20. Jul 27, 2016

### haruspex

As I posted, the vectors are not equal and opposite. They only have opposite directions. The magnitudes can be different.
How did 'and' become 'or'?