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AravindG
 4 years ago
wel i hav some doubts on 1d motion pls cm here
AravindG
 4 years ago
wel i hav some doubts on 1d motion pls cm here

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AravindG
 4 years ago
Best ResponseYou've already chosen the best response.11.how can i understand how acceleration and displacemenyt varies from a displacement time graph??

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1srry accleration and velocity i meant

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1based on that tell me hw the sign in varous intervals OA ,,AB ,BC and CD is , 0 ,+ ,0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Imagine 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
 4 years ago
Best ResponseYou've already chosen the best response.1i mean sign of acceleration

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1eashmore i asked how we can evaluate from the graph

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There 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
 4 years ago
Best ResponseYou've already chosen the best response.0My method still holds. Use the principles you learned in Calculus and the above differentials to draw conclusions.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.12.dw:1327309473228:dw

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1here its a smooth in clined planethe time taken to reach bttom is sqr(2l/g)

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1i dont understand why we can use s=1/2gt^2 directly here

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1but the answer is sqr(2l/g)

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1we just applied equaion of motion just as usual why??

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1it is said that acceleration of A and B are in ratio tan 60 to tan 30 how????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lines A and B have no acceleration. They are constant velocity plots. The second derivative of a straight line is zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If that is a displacement vs. time graph, and lines A and B are straight, then yes, I'm 100% sure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I told you why. Because the second derivative of a straight line is zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0yea i agree with that t=sqrt(2l/g sin theta)

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1but it is not there in options

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let'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:dwObserve 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
 4 years ago
Best ResponseYou've already chosen the best response.1hey i got the answer but there is no arc sin theta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\arcsin(\theta) = {1 \over \sin(\theta)}\]

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1k i got it ithought it was sin inverse

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\arcsin(\theta) = {1 \over \sin(\theta)} = \sin^{1}(\theta)\]

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.11/sin theta is cosec theta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just pretend like it is here.
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