Find the time t at which the angle between two vectors is 45

1. Jan 8, 2012

chris_0101

1. The problem statement, all variables and given/known data

Suppose a car travels around a circular track of radius 100m. Suppose the speed of the car varies in time as v = (2m/s)t. Find the time t at which the angle between the velocity vector and the acceleration vector is 45 degrees

2. Relevant equations

Position vector = $\vec{r}$ = $\hat{i}$bsin(ωt) + $\hat{j}$bcos(ωt)

3. The attempt at a solution

With the position vector, I found the velocity vector, acceleration vector and their corresponding magnitudes and they are shown below:

$\vec{v}$ = $\hat{i}$bωcos(ωt) - $\hat{j}$bωsin(ωt)
|v| = bω

$\vec{a}$ = -$\hat{i}$bω2sin(ωt) - $\hat{j}$bω2sin(ωt)
|a| = bω2

v$\bullet$a = |v||a|cosθ

And basically that is where I am stuck. I do not know how to implement the acceleration/velocity vectors and their magnitudes into the dot product equation since the dot product of v and a will equal 0.

If anyone can help me out with this one, that would be great. Thanks in advance.

2. Jan 8, 2012

Simon Bridge

Use the velocity vector as your baseline and do everything in it's frame of reference.
v points tangentially to the track.
There are two acceleration vectors - the tangential acceleration a which points along v and the centripetal acceleration ac which points perpendicularly to v

... the net acceleration will be atot=a+ac ...

So the tangent of the angle that atot makes with v is given by what?
When the angle is 45deg, what is the tangent?

3. Jan 8, 2012

chris_0101

isn't the angle that a_tot makes with v, simply 45 degrees? which is a given value

4. Jan 8, 2012

Simon Bridge

The angle depends on the speed which depends on time ... you want to know what time gets you 45degrees. What is the formula for centripetal acceleration?

5. Jan 8, 2012

chris_0101

the formula for centripetal acceleration is a_c = ((v_t)^2)/r

The centripetal acceleration is equal to:
a_c = ((2m/s)^2)/100m
a_c = 0.04 m/s^2

6. Jan 8, 2012

Simon Bridge

Is the tangential velocity a constant with time then?
I thought you said it was $v_t = (2.0m.s^{-2})t$ ?

7. Jan 8, 2012

chris_0101

yes, sorry. my mistake. I am extremely confused and lost with this question. So the tangential velocity is the magnitude of the velocity vector, so it would actually be:

a_c = (b(omega))^2 / r

8. Jan 8, 2012

Simon Bridge

You seem to be getting yourself further and further tangled up for some reason.

Lets have a recap:

in post #1 when you said the speed was varying as: v(t)=(2.0)t

isn't this the equation of the tangential speed? - yes?

9. Jan 8, 2012

chris_0101

I believe so, yes.

10. Jan 8, 2012

Simon Bridge

Cool - so if v(t)=(2.0)t, what is the magnitude of the tangential acceleration?

11. Jan 8, 2012

chris_0101

magnitude of tangential acceleration is simply the derivative of the tangential velocity, which is 2m/s^2

12. Jan 8, 2012

Simon Bridge

Good - now notice that the tangential acceleration does not change with time - this is important. Also, the tangential acceleration always points the same way as the tangential velocity[1].

The magnitude of the centripetal acceleration was correctly given by you earlier as:

$$a_c = \frac{v^2}{r}$$

That v in the numerator is the tangential speed.
Substitute the expression for tangential speed into the above to get the centripetal acceleration as a function of time.

-------------------------------------
[1] ... so long as you remember that slowing down is a negative acceleration

13. Jan 8, 2012

chris_0101

So I would get:

a_c = ((2t)^2)/R

So I have both tangential acceleration and centripetal acceleration. So I can proceed by substituting it into the equation:

tan(theta) = a_c/a_t

And I think I got it from this point now. a_t and v_t are pointing in the same direction so the above equation will provide the angle between acceleration and velocity.

Thanks, I cannot believe I over analyzed this question. You have helped me a lot, thanks for everything.

14. Jan 8, 2012

Simon Bridge

Well done.

The real cute bit is that once you have noticed that the two accelerations are the short-sides of a right angled triangle, you just ask yourself what sort of triangle has 45 degree angles... no need for trig.

If that confuses you - the tangent of 45 degrees is 1.

BTW: if you must use coordinates - pick then to exploit the symmetry of the situation ... for circular motion, pick polar coordinates. Just sayin.

Happy hacking.