Ruchi.
displacement time equation of a particle moving along x-axis is x=20+t3-12t.
a)find velocity of particle at time t=0
b)state whether the motion is uniformly acceleration or nt.
c)find position of the particle when v=0.
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apoorvk
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if that^ is the equation of the displacement, how would you get the equation for velocity?
apoorvk
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wrt 'time' ofcourse.
Aadarsh
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"Substitute" the value of "t" and find.
yakeyglee
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Hint for part (a): \(v=\frac{dx}{dt}\)
Hint for part (b): \(a=\frac{dv}{dt} = \frac{d^2x}{dt^2}\). If this is a function of time, then it is not constant.
Hint for part (c): What \(t\) values satisfies this condition?
Ruchi.
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v=dx/dt
Aadarsh
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Sorry. @yakeyglee is absolutely correct. Follow his steps and ur answer is done
Ruchi.
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as v=0,we 0=3t2-12
Ruchi.
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am i right.
Aadarsh
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Yeah, perfect @Ruchi.
Aadarsh
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So, we get|dw:1338103103383:dw|
Aadarsh
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A) answer is 12
yakeyglee
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More specifically, \(t=\pm 2\).
Ruchi.
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a)=-12m/s
Aadarsh
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But, "t" cannot be negative, so t = 2.
so, c) 2
Aadarsh
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A) -12m/s correct. sorry for my error.
Aadarsh
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On double differentiation, we get, a = 6t (I think I am correct).
Aadarsh
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So, unifrom acceleration
Aadarsh
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@Ishaan94 , @apoorvk @him1618 please help me and correct me if I am wqrong.
him1618
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Acceleration is not uniform
Aadarsh
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But how? My steps r wrong or rite? Please help me.
apoorvk
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No as a=6t, acceleration is dependent on time. So it's not uniform.
yakeyglee
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\(t\) most certainly can be negative. It simply corresponds to be fore \(t=0\).
him1618
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Bt that isnt in the domain@ yakeyglee
Aadarsh
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So @apoorvk bhai, when can we say that a is uniform? in what relation?
apoorvk
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And put in the value of 't' when v=0 from the first part, in the equation of displacement, that is put in t=2. That should give you the displacement at the time when v=0.
him1618
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When the accn is not a function of time
apoorvk
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uniform means constant - it isn't constant is it?
him1618
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In this case accn is a function of time...so its not uniform
Aadarsh
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Thanks all, I understood.
Aadarsh
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But, plz show me what will be the answer of (a), (b) and (c)
him1618
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velocity is dx/dt
yakeyglee
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@him1618 I don't see any statement of the domain...it's usually assumed \(t\in\mathbb R\), which includes negative values.
Aadarsh
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So, (a) = -12, (c) = 2, (b) = not uniform. Is this the final answer? Please say.
him1618
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it is so my man..bt not here..this is simple/...were not into time domain circuits here :P
him1618
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c=4
Aadarsh
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Is (a) and (b) what i gave the answer rite? How is (c) = 4?????????
him1618
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v=0
means
3t^2 - 12 =0
t=2
Put t=2 in x(t)
Aadarsh
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Still bhaiya, we get, 20 + 2*3 - 12*2 = 2. Am I rite????????