AravindG
  • AravindG
wel i hav some doubts on 1d motion pls cm here
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AravindG
  • AravindG
1.how can i understand how acceleration and displacemenyt varies from a displacement time graph??
AravindG
  • AravindG
srry accleration and velocity i meant
AravindG
  • AravindG
based on that tell me hw the sign in varous intervals OA ,,AB ,BC and CD is -, 0 ,+ ,0

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AravindG
  • AravindG
|dw:1327309166279:dw|
anonymous
  • anonymous
Imagine a simple case of a parabolic x vs. t graph. We find that the equation of the line on the graph is\[x = 10 t^2\]As would be the case in constant acceleration. Now, we know that velocity is defined as the change in distance over the change in time. In differential form this is\[v = {dx \over dt}\]And additionally, that acceleration is the change in velocity over the change in time. Again, in differential form, this is\[a = {dv \over dt} = {d^2 x \over dt^2}\] It can readily be seen, that to find the velocity of the above displacement equation, we simply differentiate with respect to time. \[v = 20 t\]and then differential the velocity with respect to time to find the acceleration.\[a = 20\]
AravindG
  • AravindG
i mean sign of acceleration
AravindG
  • AravindG
eashmore i asked how we can evaluate from the graph
AravindG
  • AravindG
look the graph
anonymous
  • anonymous
There are infinite possibilities to your second question. But since we have observed that \[a = {d^2 x \over dt^2}\]it is apparent that acceleration is the second derivative of the x vs. t curve.
anonymous
  • anonymous
My method still holds. Use the principles you learned in Calculus and the above differentials to draw conclusions.
AravindG
  • AravindG
2.|dw:1327309473228:dw|
AravindG
  • AravindG
here its a smooth in clined planethe time taken to reach bttom is sqr(2l/g)
AravindG
  • AravindG
i dont understand why we can use s=1/2gt^2 directly here
AravindG
  • AravindG
ipls help
AravindG
  • AravindG
but the answer is sqr(2l/g)
AravindG
  • AravindG
we just applied equaion of motion just as usual why??
AravindG
  • AravindG
|dw:1327309719031:dw|
AravindG
  • AravindG
it is said that acceleration of A and B are in ratio tan 60 to tan 30 how????
AravindG
  • AravindG
helo pls help
anonymous
  • anonymous
Patience....
AravindG
  • AravindG
k
anonymous
  • anonymous
Lines A and B have no acceleration. They are constant velocity plots. The second derivative of a straight line is zero.
AravindG
  • AravindG
are u sure?/
anonymous
  • anonymous
If that is a displacement vs. time graph, and lines A and B are straight, then yes, I'm 100% sure.
AravindG
  • AravindG
y is a=0??
anonymous
  • anonymous
I told you why. Because the second derivative of a straight line is zero.
AravindG
  • AravindG
k tel me the othr one
anonymous
  • anonymous
I don't think it is right either. The time it takes to reach the bottom will be\[T = \sqrt{2l \over g \sin(\theta)}\]
anonymous
  • anonymous
yea i agree with that t=sqrt(2l/g sin theta)
AravindG
  • AravindG
but it is not there in options
AravindG
  • AravindG
1 Attachment
anonymous
  • anonymous
Let's analyze the equations of motion here. The only force acting on the particle moving down the incline is that of gravity. Therefore, we can say with certainty that\[\vec a = -g\]From the definition of acceleration, we can also say that\[\vec a = { d \vec v \over dt}\]Separating variables\[d \vec v = \vec a dt\]Integrating\[\int\limits d \vec v = \int\limits \vec a dt \rightarrow \vec v = \vec a *t\]By the same token, we can observe that\[\vec s = \vec v_0*t+ {1 \over 2} \vec a *t^2\]Take careful note of the fact that displacement, velocity, and acceleration are vector quantities. Now, let's analyze the kinematics of the problem. Let's fix a frame such that \(\hat i\) is directed from the top of the incline to the bottom and parallel to the inclined surface, \(\hat j\) is directed away from the inclined surface and perpendicular to the inclined surface, and \(\hat k\) is \(\hat i \times \hat j\). We will also fix a ground, inertial frame with unit vectors \((\hat I,\hat J,\hat K)\)|dw:1327390446864:dw|Observe that the direction cosines are \[\left(\begin{matrix}\hat i = \cos(\theta) \hat I - \sin(\theta) \hat J \\ \hat j = \sin(\theta) \hat I + \cos(\theta) \hat J\end{matrix}\right)\]Let's also observe that\[\vec a = -g \hat J\]and, from the above direction cosines, that\[\vec a = g \sin(\theta) \hat i\]The particle will move along the inclined plane with a displacement of \[\vec s = l \hat i\]From the equation I gave earlier for displacement, we can observe that\[l \hat i = \vec v_0 \hat i * t + {1 \over 2} g \sin(\theta) \hat i * t^2\]Simplifying by observing that \(v_0 = 0\), and that \(\hat i\) cancels out through all terms, we get\[l = {1 \over 2} g \sin(\theta) \cdot t^2\]Solving for \(t\)\[t = \sqrt{ 2l \over g \sin(\theta)}\]This isn't an answer choice, so let's see if we can manipulate it to match one of the answers. Observe from trigonometry that\[\sin(\theta) = \frac{h}{l}\]Solving for \(l\)\[l = {h \over \sin(\theta)}\]Let's substitute this into our expression for \(t\), yielding\[t = \sqrt{2 h \over g \sin^2(\theta)}\]Removing \(\sin(\theta)\) from the radical yields, \[t = \arcsin(\theta) \sqrt{2 h \over g}\]which appears to be answer choice c.
AravindG
  • AravindG
hey i got the answer but there is no arc sin theta
anonymous
  • anonymous
\[\arcsin(\theta) = {1 \over \sin(\theta)}\]
AravindG
  • AravindG
lol
AravindG
  • AravindG
k i got it ithought it was sin inverse
anonymous
  • anonymous
\[\arcsin(\theta) = {1 \over \sin(\theta)} = \sin^{-1}(\theta)\]
AravindG
  • AravindG
hey thts not
AravindG
  • AravindG
1/sin theta is cosec theta
anonymous
  • anonymous
Yeah. You're right.
anonymous
  • anonymous
Just pretend like it is here.
AravindG
  • AravindG
:)

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