- #1
SqueeSpleen
- 141
- 5
Hi. Sorry, couldn't make a more specific tittle, I'm not used to solve physics problems.
A friend of mine handed me the following problem:
You have a graph that denotes the aceleration of an object as a function of the position of the object. This object has a mass of 10 Kg.
What's the speed when the position is equal to 4 meters?
$$ r(t) = a \dfrac{t^{2}}{2} + v_{0}.t $$
$$ F=m.a $$[/B]
What I consider I did right:
I have translated the problem to the following one, I guess it's right but it's too complicated:
You have a object that's initially still, and then you accelerate it at 6m/s^2 for certain time t, until it goes 2 meters. Then, you the acceleration is linear and I wrote it as 6m/s^2 times (4m-r(t))/2m times 6m/s^2 where r(t) denotes the position as function of the time.
First step to solve it, as the acceleration is constant in the beggining:
$$ 6 \frac{m}{s^{2}}.\dfrac{t^{2}}{2} = 2m $$
$$ t^{2} = \frac{2}{3} s^{2} $$
$$ t = \sqrt{ \frac{2}{3} } s = \dfrac{\sqrt{6}}{3}$$
Also using this we know that the speed of the object when it's at 2 meters of it's original location
$$ \sqrt{ \frac{2}{3} } s 6 \frac{m}{s^{2}}=2. \sqrt{6} m/s $$
Where I think I got things messy:
So I make the acceleration as function of time, we have
$$a(t) = \frac{4m-r(t)}{2m} \cdot 6 \frac{m}{s^{2}} $$
$$a(t) = 12\frac{m}{s^{2}} 3-r(t) \frac{1}{s^{2}} $$
Drop the units and remplace a(t) for r''(t)
$$r''(t)+3r(t) = 12$$
$$r( \dfrac{\sqrt{6}}{3} ) =2 \qquad r'(\dfrac{\sqrt{6}}{3})=\dfrac{\sqrt{6}}{3}.6=2 \sqrt{6} $$
Solved the initial value problem, the solution which I think is valid when r(t) is between 2 and 4 is:
$$ 2 \cdot \left( \sqrt{2} . \sin \left( \sqrt{2} - \sqrt{3} \cdot t \right) +cos( \sqrt{2} -\sqrt{3} \cdot t ) -2 ) \right) $$
Put r(t)=4 and got
$$t_{4} = \dfrac{ \left( arctan \left( \dfrac{ \sqrt{2} }{2} \right) + \sqrt{2} \right) \sqrt{3} }{3} $$
So r'(t_{4}) is 6m/s, but I'm sure there's a better way to solve it, but I have too little physics training to find out by myself.
Any ideas?
A friend of mine handed me the following problem:
Homework Statement
You have a graph that denotes the aceleration of an object as a function of the position of the object. This object has a mass of 10 Kg.
What's the speed when the position is equal to 4 meters?
Homework Equations
$$ r(t) = a \dfrac{t^{2}}{2} + v_{0}.t $$
$$ F=m.a $$[/B]
The Attempt at a Solution
What I consider I did right:
I have translated the problem to the following one, I guess it's right but it's too complicated:
You have a object that's initially still, and then you accelerate it at 6m/s^2 for certain time t, until it goes 2 meters. Then, you the acceleration is linear and I wrote it as 6m/s^2 times (4m-r(t))/2m times 6m/s^2 where r(t) denotes the position as function of the time.
First step to solve it, as the acceleration is constant in the beggining:
$$ 6 \frac{m}{s^{2}}.\dfrac{t^{2}}{2} = 2m $$
$$ t^{2} = \frac{2}{3} s^{2} $$
$$ t = \sqrt{ \frac{2}{3} } s = \dfrac{\sqrt{6}}{3}$$
Also using this we know that the speed of the object when it's at 2 meters of it's original location
$$ \sqrt{ \frac{2}{3} } s 6 \frac{m}{s^{2}}=2. \sqrt{6} m/s $$
Where I think I got things messy:
So I make the acceleration as function of time, we have
$$a(t) = \frac{4m-r(t)}{2m} \cdot 6 \frac{m}{s^{2}} $$
$$a(t) = 12\frac{m}{s^{2}} 3-r(t) \frac{1}{s^{2}} $$
Drop the units and remplace a(t) for r''(t)
$$r''(t)+3r(t) = 12$$
$$r( \dfrac{\sqrt{6}}{3} ) =2 \qquad r'(\dfrac{\sqrt{6}}{3})=\dfrac{\sqrt{6}}{3}.6=2 \sqrt{6} $$
Solved the initial value problem, the solution which I think is valid when r(t) is between 2 and 4 is:
$$ 2 \cdot \left( \sqrt{2} . \sin \left( \sqrt{2} - \sqrt{3} \cdot t \right) +cos( \sqrt{2} -\sqrt{3} \cdot t ) -2 ) \right) $$
Put r(t)=4 and got
$$t_{4} = \dfrac{ \left( arctan \left( \dfrac{ \sqrt{2} }{2} \right) + \sqrt{2} \right) \sqrt{3} }{3} $$
So r'(t_{4}) is 6m/s, but I'm sure there's a better way to solve it, but I have too little physics training to find out by myself.
Any ideas?
Attachments
Last edited: