# Could you help me review and check my physics homework?

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 12th 2012, 05:07 AM
Glyper
Could you help me review and check my physics homework?
So I have four tasks to do and I have a guess for some of them but would greatly appreciate if you could help me be sure of it or clarify what I don't understand.

1. A car covers 2000m in 2 minutes and 20 seconds. The maximum velocity is 60kmph. In the beginning and in the end of its move, the car moves with the same acceleration. Determine the acceleration.
2. What are the angular velocities of the three clock hands?
3. What is the angular velocity of a car's wheel driving 54kmph if its diameter equals 150cm?
4. We have a liquid in a test-tube moving on a circle of diameter 10cm. To find the element we want to determine, we have to cause an acceleration of 100000g. How many rounds per minute does the circle have to make?

So here is what I think:

1. I'm assuming it started with 0kmph and eventually gained the said 60kmph - is it a good way to go? If so, then: $\displaystyle a=\frac{60kmph}{140s}\approx\frac{16,7m/s}{140s}=0,1192\frac{m}{s^2}$
2. $\displaystyle \omega_m=\frac{2\pi}{3600s} \approx \frac{6,28}{3600s} \approx 0,00174rad/s$
$\displaystyle \omega_h=\frac{2\pi}{43200s}\approx\frac{6,28}{432 00s}\approx0,0001453rad/s$
$\displaystyle \omega_s=\frac{2\pi}{60s}\approx\frac{6,28}{60s} \approx 0,1rad/s$
3. The diameter is 1,5m so the circumference is 9,42m. It moves 54kmph so it's 54000m/h which is 15m/s. 15m is 1,592 of the circumference so $\displaystyle \omega=\frac{1,592\cdot2\pi}{1s} \approx 10rad/s$.
4. How should I approach how to "cause an acceleration of 100000g? And why is acceleration here measured in grams?

Could you please check what I did and help me with what I did wrong or couldn't do? Thank you very much in advance.
• Oct 12th 2012, 07:20 AM
ebaines
Re: Could you help me review and check my physics homework?
Number 1. is incorrect. Calculate the distance traveled while accelerating at 0.1192 m/s^2 and you'll see the car won't cover 2000 meters in 140 seconds. The acceleration needs to be greater, so that the car reaches 60 KPH and cruises at that speed for a certain distance, then decelerates at the end. If we assume that the car comes to a stop at exactly 2000 meters in 140 seconds you can set up two equations like this:

$\displaystyle 2000m = \frac 1 2 at_1^2 + v_1t_2 + (v_1t_1-\frac 1 2 a t_1^2)$

$\displaystyle 2 t_1 + t_2 = 140s$

Where v_1 = 60 KPH. Solve for t_1, and then determine a = v_1/t_1

2 is correct

3 is off by a factor ten - the wheel diameter is 0.15m, not 1.5m.

4. The term "g" here means "acceleration of gravity," not grams. The acceleration due to gravity is 9.8m/s^2. So they're asking what the rotation speed must be to generate centripedal acceleration equal to 100,000 x 9.8m/s^2 .
• Oct 12th 2012, 08:44 AM
Krahl
Re: Could you help me review and check my physics homework?
Quote:

Originally Posted by ebaines
3 is off by a factor ten - the wheel diameter is 0.15m, not 1.5m.

There are 100 centimeters in a meter.

1. Driving at constant acceleration to reach 2000m gives the equation
$\displaystyle S = 2000 = at^2+v_0 t + S_0$

There has been nothing said about the initial velocity so we cannot assume it to be 0. $\displaystyle S_0=0$ since we are covering 2000 meters from 0 to 2000.
Another clue is that the maximum velocity is $\displaystyle 60kmph=\frac{50}{3}mps$. So

$\displaystyle v_{max} = at + v_0=\frac{50}{3}$ so that
$\displaystyle v_0=\frac{50}{3}-at$

Replacing $\displaystyle v_0$ into the original equation gives
$\displaystyle S = 2000 = at^2+(\frac{50}{3}-at) t$

We know the $\displaystyle t=140s$ so now you can solve this to find a. You can then solve $\displaystyle v_0$ if you wanted to.
• Oct 12th 2012, 09:01 AM
ebaines
Re: Could you help me review and check my physics homework?
Quote:

Originally Posted by Krahl
There are 100 centimeters in a meter.

Ah... right. Thanks for catching this. I'm afraid I was thinking mm. OK then, number 3 is correct!
• Oct 12th 2012, 09:15 AM
Glyper
Re: Could you help me review and check my physics homework?
Thank you ebaines. I have a question regarding 1., though. Namely - how did you come up with the first equation?

Quote:

Originally Posted by ebaines
$\displaystyle 2200m = \frac 1 2 at_1^2 + v_1t_2 + (v_1t_1-\frac 1 2 a t_1^2)$
$\displaystyle 2 t_1 + t_2 = 140s$

I understand that $\displaystyle t_1$ is the time it takes the train to accelerate to full speed and to brake to zero from the max. However, how did the first line come up? I know the generalised formula for accelerated motion but it's nothing alike...

And for fourth - thank you for clarification. So I'd have something like this: $\displaystyle 10^6 \cdot 9.8\frac{m}{s^2}=\frac{\delta v}{\delta t}$. It seems that whatever $\displaystyle \delta v$ I choose, I'll have to make it such that when divided by the denominator, it equals left side. So is the answer just... $\displaystyle 10^6 \cdot 9.8\frac{m}{s}$?

EDIT: Krahl, thank you very much too! I believe we can assume the initial velocity to be 0, though. The tasks I wrote from my head, not word-to-word from paper and it turns out there's an information that the car "cover the distance between two stops" in the said time. Then, I think it may be OK to assume the $\displaystyle v_0=0$, shouldn't it?
• Oct 12th 2012, 09:34 AM
Krahl
Re: Could you help me review and check my physics homework?
@Glyper

The question is a little ambiguous. If we are to assume there is a constant acceleration throughout the journey the only initial velocity is 11.9 m/s given by the equation I've written. The only way we can accept a zero initial velocity is if we interpret the sentence "In the beginning and in the end of its move, the car moves with the same acceleration" as meaning that the car can alter acceleration but must be equal at the two ends. In this case there isn't a single solution since the car could accelerate from 0kmph to 60kmph in less than a second drive all the way to just before the 2000m line taking around 120 seconds, stop then accelerate at the 140s mark to finish it off.
• Oct 12th 2012, 09:45 AM
ebaines
Re: Could you help me review and check my physics homework?
I interpret the phrase form the question as written: "In the beginning and in the end of its move, the car moves with the same acceleration" to imply that there are three phases to the car's motion: (1) phase 1 where it accelerates to 60 KPH, but we don't know if it has an initial velocity or not, (2) phase 2 where it moves at a constant 60 KPH, and (3) phase 3 where it decelerates to 0, with acceleration equal to the negative of phase 1. It turns out that there are too many unknowns to solve this if the inital velocity is not given, so I assume it's some value. I assume the initial and final velocities are both 0, because otherwise the problem can't be solved. The distance travelled in each phase can be found from $\displaystyle d_i = v_0t + \frac 1 2 a t^2$. For phase 1 this is $\displaystyle d_1 = \frac 1 2 a t_1^2$. For phase 2: $\displaystyle d_2 = v_1 t_2$, and for phase 3: $\displaystyle d_3 = v_1t - \frac 1 2 a t_3^2$. Since $\displaystyle t_1 = t_3$, you can add these up to get the total distance travelled:

$\displaystyle d = 2000 m = d_1 + d_2 + d_3 = \frac 1 2 at_1^2 + v_1t_2 + (v_1t_1- \frac 1 2 a t_1^2)$
• Oct 12th 2012, 09:50 AM
Krahl
Re: Could you help me review and check my physics homework?
Acceleration and deceleration are not the same however. They are negatives of each other. It is possible the original question stated that the magnitudes are equal.
• Oct 12th 2012, 10:01 AM
ebaines
Re: Could you help me review and check my physics homework?
Quote:

Originally Posted by Krahl
Acceleration and deceleration are not the same however. They are negatives of each other. It is possible the original question stated that the magnitudes are equal.

Agreed - which is why I noted the assumption that the deceleration is equal to the negative of the value of acceleration in phase 1. As you said earlier - the question is ambiguous. Since Glyper was just paraphrasing the question as originally presented to him, only he knows if all my assumptions are good or not.
• Oct 14th 2012, 04:52 AM
Glyper
Re: Could you help me review and check my physics homework?
OK, I got my hands on the original question. It goes as follows (literally):

A car covers 2000 meters between two traffic lights at both of which it has to stop in 2 minutes and 20 seconds. The maximum velocity of the car is 60kmph. In the beginning and in the end of its move, the car drives with the same acceleration. To determine the acceleration.

So for my thoughts, it means that in the beggining, the velocity is 0. So it is in the end. It surely does feel a little ambiguous, but that's the literal task. Which of the formulas you both have given should work here, then?

Quote:

And for fourth - thank you for clarification. So I'd have something like this: $\displaystyle 10^6 \cdot 9.8\frac{m}{s^2}=\frac{\delta v}{\delta t}$. It seems that whatever $\displaystyle \delta v$ I choose, I'll have to make it such that when divided by the denominator, it equals left side. So is the answer just... $\displaystyle 10^6 \cdot 9.8\frac{m}{s}$?
• Oct 15th 2012, 05:00 AM
ebaines
Re: Could you help me review and check my physics homework?
Quote:

Originally Posted by Glyper
Which of the formulas you both have given should work here, then?

We both gave explanations of the assumptions we used - so which do you think corresponds to the question as written?

Quote:

Originally Posted by Glyper

As I pointed out earlier, you need to find the rotational velocity that yields an acceleration of 100,000 g's. To get you started you should know this formula for centripetal acceleration:

$\displaystyle a = \omega^2 R$

where a = acceleration, which is 100,000 x 9.8 m/s^2, R = 0.05m, and so solve for $\displaystyle \omega$, which is the rotational velocity in radians per second. Then convert that to revolutions per minute.
• Oct 15th 2012, 09:06 AM
Glyper
Re: Could you help me review and check my physics homework?
Thank you. I'll think about the fourth, then, so let's focus back on the first: I supposed yours corresponds to the question. However, I can't get it to work - namely, we start with:

$\displaystyle 2000m = \frac 1 2 at_1^2 + v_1t_2 + (v_1t_1-\frac 1 2 a t_1^2)$
$\displaystyle 2 t_1 + t_2 = 140s$

In the first equation, the $\displaystyle \frac{at^2}{2}$ cancel out and we have $\displaystyle 2000m=60(t_1+t_2)$. This, however, is clearly contradictory with the second line as t1 would have to be 106 seconds and it can't. Where do I go wrong?
• Oct 15th 2012, 09:24 AM
ebaines
Re: Could you help me review and check my physics homework?
Quote:

Originally Posted by Glyper
Where do I go wrong?

Watch your units. In the second equation you're equating meters on the left with kilometers per hour times seconds on the right. I suggest you change 60 KPH to the equivalent in meters per second and try again.
• Oct 15th 2012, 09:59 AM
Glyper
Re: Could you help me review and check my physics homework?
Right you are! It turns out the acceleration would be $\displaystyle 0,835\frac{m}{s^2}$ and it definitely feels right :) Thank you.

And for the fourth - thank you a lot for the formula. Then, it looks like we have 3,16rads per second which is 30rpm, right?
• Oct 15th 2012, 10:42 AM
ebaines
Re: Could you help me review and check my physics homework?
Quote:

Originally Posted by Glyper
Right you are! It turns out the acceleration would be $\displaystyle 0,835\frac{m}{s^2}$ and it definitely feels right :) Thank you.

You're welcome!

Quote:

Originally Posted by Glyper
And for the fourth - thank you a lot for the formula. Then, it looks like we have 3,16rads per second which is 30rpm, right?

Two issues:

1. I think you used R=0.1m for the radius of the circle, but the diameter of the circle is 10 cm, which means its radius is 5cm.
2. I think you calculated rotational velocity to give an acceleration of just 1 g, but the required acceleration is 100,000 times g.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last