Find time at which position vector points towards origin

[Taniaz]

Homework Statement

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.[/B]

Homework Equations

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

[/B]

 

[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?

 

[Taniaz]

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

 

[haruspex]

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

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

 

[Taniaz]

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

 

[haruspex]

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

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.

 

[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?

 

[haruspex]

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?

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

 

[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?

 

[haruspex]

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?

Yes.

 

[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?

 

[haruspex]

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?

Yes, but only the direction, not the magnitude

 

[Taniaz]

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

 

[haruspex]

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

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

 

[Taniaz]

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

 

[haruspex]

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

Two definite values.

 

[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

 

[haruspex]

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

Please post your working.

 

[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

 

[haruspex]

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

As I posted, the vectors are not equal and opposite. They only have opposite directions. The magnitudes can be different.

(-2t)i = (-1+t^2)i and (3-10t)j=(-3t+5t^2)j
t = -1-sqrt2 or t = sqrt 2-1 f

How did 'and' become 'or'?

 

[Taniaz]

Sorry I meant and. Is this correct though?

 

[RUber]

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

You are looking for one t that satisfies both the i and j components at the same time.
Let v = cR, where c is any non-zero constant. If c is negative then the velocity is pointing toward the origin. When c is positive, the velocity is pointing away from the origin.

 

[Taniaz]

Ah that makes sense! Thank you Haruspex and RUber!

 

[RUber]

After adding in the c, you should be able to find a solution that satisfies your and condition.
$t = ( -1 - \sqrt{2} \text{ or } -1 + \sqrt{2} ) \text{ and } ( \frac{-7 - \sqrt{109}}{10} \text{ or } \frac{-7 + \sqrt{109}}{10} )$ are inconsistent.
$-2t \hat i + (3-10t) \hat j = (c-ct^2 )\hat i - (3ct + 5ct^2)\hat j \\ ct^2 -2t -c = 0 \text{ and } 5ct^2 + (3c-10)t + 3 = 0$

 

[Taniaz]

Correct me if I've misunderstood what you just said. So up till the values of t, it's correct? We just need to find which values of t give c as negative and that will indicate that it's pointing to the origin?

 

[RUber]

The work you posted in #19 is correct for v = (-1)R. Replace -1 with a c, and rewrite the equations.

 

[Taniaz]

I understand but then how do you solve for c?
ct^2−2t−c=0 and 5ct^2+(3c−10)t+3=0
I put in the values for t that I found and then see if c is negative?

 

[Taniaz]

Ok so the values of t all seem to give c=-1 which seems to be alright in this case. I'm guessing this equations are general ct^2−2t−c=0 and 5ct^2+(3c−10)t+3=0 where you can substitute for whatever you think c is and solve for t?

 

[RUber]

I don't see how c = -1 works.
You have two times for i and two different times for j.
You need a time that satisfies both i and j.

The two equations in #24 must both be true. One method is to solve for t in terms of c using linear algebra, then solve one or both of the equations for t as a quadratic.
Finally set your solutions for t equal to each other to find c. You should be able to find a positive and a negative c.

 

[Taniaz]

I understand what you're saying but isn't this a little too complicated for a 1 mark A'Level question?

 

[ehild]

I understand what you're saying but isn't this a little too complicated for a 1 mark A'Level question?

You are looking for one t that satisfies both the i and j components at the same time.
Let v = cR, where c is any non-zero constant. If c is negative then the velocity is pointing toward the origin. When c is positive, the velocity is pointing away from the origin.

$\vec v = c \vec R$ means that v_x=c R_x and v_y =c R_y. Dividing the first equation with the second one, v_x/v_y=R_x/Ry . Substitute the expressions in terms of t,
\frac{1-t^2}{3t-5t^2}=\frac{-2t}{3-10t}
and solve for t. You get two solutions. One of them means a positive c , the other solution means a negative one.

 

[Taniaz]

Thank you RUber, Haruspex and ehild for your help! Much appreciated! :)

 

[haruspex]

Thank you RUber, Haruspex and ehild for your help! Much appreciated! :)

If you are familiar with cross products, another route is to note that the collinearity of the position and velocity vectors means their cross product is zero. This avoids bringing in c and gives you an equation in t only (though it might be nasty). Having extracted the possible values of t, you could then evaluate the two vectors at those times to find out which has velocity towards the origin and which away.

 

[Taniaz]

So I got that t= 1/3 and t = 3
At t=3, c = 3/4 and at t=1/3, c = -3/4
So at t=1/3 it's pointing towards the origin.

 

[haruspex]

So I got that t= 1/3 and t = 3
At t=3, c = 3/4 and at t=1/3, c = -3/4
So at t=1/3 it's pointing towards the origin.

Looks right.