wel i hav some doubts on 1d motion pls cm here

- AravindG

wel i hav some doubts on 1d motion pls cm here

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- AravindG

1.how can i understand how acceleration and displacemenyt varies from a displacement time graph??

- AravindG

srry accleration and velocity i meant

- AravindG

based on that tell me hw the sign in varous intervals OA ,,AB ,BC and CD is -, 0 ,+ ,0

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## More answers

- AravindG

|dw:1327309166279:dw|

- 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

i mean sign of acceleration

- AravindG

eashmore i asked how we can evaluate from the graph

- AravindG

look the graph

- 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

My method still holds. Use the principles you learned in Calculus and the above differentials to draw conclusions.

- AravindG

2.|dw:1327309473228:dw|

- AravindG

here its a smooth in clined planethe time taken to reach bttom is sqr(2l/g)

- AravindG

i dont understand why we can use s=1/2gt^2 directly here

- AravindG

ipls help

- AravindG

but the answer is sqr(2l/g)

- AravindG

we just applied equaion of motion just as usual why??

- AravindG

|dw:1327309719031:dw|

- AravindG

it is said that acceleration of A and B are in ratio tan 60 to tan 30 how????

- AravindG

helo pls help

- anonymous

Patience....

- AravindG

k

- anonymous

Lines A and B have no acceleration. They are constant velocity plots. The second derivative of a straight line is zero.

- AravindG

are u sure?/

- anonymous

If that is a displacement vs. time graph, and lines A and B are straight, then yes, I'm 100% sure.

- AravindG

y is a=0??

- anonymous

I told you why. Because the second derivative of a straight line is zero.

- AravindG

k tel me the othr one

- 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

yea i agree with that t=sqrt(2l/g sin theta)

- AravindG

but it is not there in options

- AravindG

##### 1 Attachment

- 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

hey i got the answer but there is no arc sin theta

- anonymous

\[\arcsin(\theta) = {1 \over \sin(\theta)}\]

- AravindG

lol

- AravindG

k i got it ithought it was sin inverse

- anonymous

\[\arcsin(\theta) = {1 \over \sin(\theta)} = \sin^{-1}(\theta)\]

- AravindG

hey thts not

- AravindG

1/sin theta is cosec theta

- anonymous

Yeah. You're right.

- anonymous

Just pretend like it is here.

- AravindG

:)

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