Magnitude of velocity and acceleration

[JustDerek]

A car travels around a circular track with a radius of r=250m. When it is at point A then V_A=5m/s which increases at a rate of a=(0.06t)m/s². Determine the magnitude of its velocity and its acceleration when it is 1/3 around the track. My distance in this situation is obviously 524m.

Can anyone help please? I've seen numerous different answers but none of these are what I need. I've tried integrating it twice but this gives the value of t as a complex number which doesn't really help me.
Thanks in advance
Derek

 

[JustDerek]

Been informed that no integration is needed but this leaves me lost as to how to solve the problem.

 

[Randy Beikmann]

A car travels around a circular track with a radius of r=250m. When it is at point A then V_A=5m/s which increases at a rate of a=(0.06t)m/s². Determine the magnitude of its velocity and its acceleration when it is 1/3 around the track. My distance in this situation is obviously 524m.

Can anyone help please? I've seen numerous different answers but none of these are what I need. I've tried integrating it twice but this gives the value of t as a complex number which doesn't really help me.
Thanks in advance
Derek

I think you're making this too hard. Without doing the problem for you, I'll say that you've been given an initial tangential velocity, a constant tangential acceleration, and a tangential distance. You can calculate the final tangential velocity from this. Now, you already knew the final tangential acceleration, and you can now calculate the final radial acceleration from the tangential velocity. Then find the vector magnitude of that acceleration vector.

 

[JustDerek]

Hey thanks for looking at this. I'll have a look using the hints you gave and let you know the result.

 

[JustDerek]

The main thing confusing me is that the acceleration is $0.06tm/s^2$ but I don't know the time and I'm unsure how to calculate it

 

[PeroK]

The main thing confusing me is that the acceleration is $0.06tm/s^2$ but I don't know the time and I'm unsure how to calculate it

How would you solve the problem if the track were straight?

 

[JustDerek]

How would you solve the problem if the track were straight?

Isn't the magnitude in both directions though, the tangential and normal directions, which is why the question is a curve?

 

[PeroK]

Isn't the magnitude in both directions though, the tangential and normal directions, which is why the question is a curve?

I assume you mean the magnitude of the "acceleration". You should read the question more carefully. Where does it say acceleration?

 

[JustDerek]

I assume you mean the magnitude of the "acceleration". You should read the question more carefully. Where does it say acceleration?

It's in two parts. First it says to find the magnitude of the velocity at 1/3 of the way round the track. Once that is found it asks to find the magnitude of acceleration for the same distance. It actually gives acceleration as v with the dot above it so it gives $\dot{v}=0.06t^2$

 

[PeroK]

It's in two parts. First it says to find the magnitude of the velocity at 1/3 of the way round the track. Once that is found it asks to find the magnitude of acceleration for the same distance. It actually gives acceleration as v with the dot above it so it gives $\dot{v}=0.06t^2$

Are you sure that's the question? How do you know that $v$ is the velocity vector and not the speed?

 

[PeroK]

. When it is at point A then V_A=5m/s which increases at a rate of a=(0.06t)m/s².

That says to me that $V_A$ is the speed, which increases at a rate of $a$.

Velocity, strictly speaking, doesn't increase, it changes. If $a$ were the magnitude of the acceleration, I would expect the question to say so.

 

[JustDerek]

That says to me that $V_A$ is the speed, which increases at a rate of $a$.

Velocity, strictly speaking, doesn't increase, it changes. If $a$ were the magnitude of the acceleration, I would expect the question to say so.

That's the question. You can't see the ##\dot{v} on there but on the desktop version I assure you that you can

 

[PeroK]

That's the question. You can't see the ##\dot{v} on there but on the desktop version I assure you that you can

It's poorly worded. It should say speed, rather than velocity. In any case, if you really are given the magnitude of the acceleration as a constant, then Part B would be rather easy!

 

[JustDerek]

It's poorly worded. It should say speed, rather than velocity. In any case, if you really are given the magnitude of the acceleration as a constant, then Part B would be rather easy!

But I need to find t I presume and then I can find the magnitude of velocity, but I'm not sure how to find t. The whole question is confusing.

 

[PeroK]

But I need to find t I presume and then I can find the magnitude of velocity, but I'm not sure how to find t. The whole question is confusing.

Assume my interpretation. You know the speed, you know the increase in speed, you know the distance ...

 

[JustDerek]

Assume my interpretation. You know the speed, you know the increase in speed, you know the distance ...

I'm still struggling but I don't want to try your patience anymore and I appreciate you taking the time to help greatly.

 

[PeroK]

I'm still struggling but I don't want to try your patience anymore and I appreciate you taking the time to help greatly.

If you understand 1D kinematics, can you relate that to motion in a circle (or any other curve) if you are given the speed along the curve, the increase in speed along the curve and the distance along the curve?

 

[JustDerek]

If you understand 1D kinematics, can you relate that to motion in a circle (or any other curve) if you are given the speed along the curve, the increase in speed along the curve and the distance along the curve?

But all I know is that the increase in speed is $0.06*t$ surely I need to know t before o can find the rest? Sorry I'm doing this degree after an 8 year break since I did the first year of dynamics. Sometimes a question pops up I just can't get my head around and then I have the "oh of course it is" moment ages later

 

[jbriggs444]

Point A appears to be the starting point for the exercise. A reasonable assumption is that t=0 when the car passes point A.

 

[PeroK]

But all I know is that the increase in speed is $0.06*t$ surely I need to know t before o can find the rest? Sorry I'm doing this degree after an 8 year break since I did the first year of dynamics. Sometimes a question pops up I just can't get my head around and then I have the "oh of course it is" moment ages later

So, you have a variable rate of increase in speed. Do you know how to handle that in 1D?

 

[JustDerek]

So, you have a variable rate of increase in speed. Do you know how to handle that in 1D?

I'm not sure which formula to use to be honest. I know there's three for kinematics in one dimension but I don't know which to use.

 

[PeroK]

Been informed that no integration is needed but this leaves me lost as to how to solve the problem.

Who said you don't have to integrate?

I'm not sure which formula to use to be honest. I know there's three for kinematics in one dimension but I don't know which to use.

There's no "formula" for variable acceleration. You have to integrate!

 

[JustDerek]

Who said you don't have to integrate?
There's no "formula" for variable acceleration. You have to integrate!

I've integrated twice. I got $v=0.03t^2+5t$ and then integrated again to get $s=0.01^3+2.5t^2+c$ but even here I'm unsure how to isolate the t's because they have different powers and values attached to them

 

[PeroK]

I've integrated twice. I got $v=0.03t^2+5t$ and then integrated again to get $s=0.01^3+2.5t^2+c$ but even here I'm unsure how to isolate the t's because they have different powers and values attached to them

You made a mistake somewhere. Numbers hide the physics, so you'll never spot a mistake with numbers. Try using $a$ and $v_o$ and then you can sanity check any formula.

 

[JustDerek]

I've integrated twice. I got $v=0.03t^2+5t$ and then integrated again to get $s=0.01^3+2.5t^2+c$ but even here I'm unsure how to isolate the t's because they have different powers and values attached to them

also i know this gives $524=0.01^3+2.5t^2+c$

 

[JustDerek]

You made a mistake somewhere. Numbers hide the physics, so you'll never spot a mistake with numbers. Try using $a$ and $v_o$ and then you can sanity check any formula.

ok now my mind is just gone, I'm going to leave it for a while and come back to it another time

 

[PeroK]

also i know this gives $524=0.01^3+2.5t^2+c$

Let me help you a bit more. The formula you got was:

$d(t) = \frac{at^3}{6} + \frac{v_0 t^2}{2}$

Now, because it's algebra, rather than meaningless numbers, you can try $a=0$ and see whether that makes sense.

 

[jbriggs444]

also i know this gives $524=0.01^3+2.5t^2+c$

You mean $524=0.01t^3 + 2.5t^2 + c$ where t = the ending time.
But you also know a boundary condition, $0 = 0.01t^3 + 2.5t^2 + c$ where t = the starting time. Solve that for c (easy peasy).

 

[JustDerek]

Let me help you a bit more. The formula you got was:

$d(t) = \frac{at^3}{6} + \frac{v_0 t^2}{2}$

Now, because it's algebra, rather than meaningless numbers, you can try $a=0$ and see whether that makes sense.

You right I did integrate it wrong it's $5t$ and not $2.5t^2$

 

[JustDerek]

Wouldn't that make c=0 at the start point?