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.

- anonymous

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- apoorvk

if that^ is the equation of the displacement, how would you get the equation for velocity?

- apoorvk

wrt 'time' ofcourse.

- anonymous

"Substitute" the value of "t" and find.

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## More answers

- anonymous

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?

- anonymous

v=dx/dt

- anonymous

Sorry. @yakeyglee is absolutely correct. Follow his steps and ur answer is done

- anonymous

as v=0,we 0=3t2-12

- anonymous

am i right.

- anonymous

Yeah, perfect @Ruchi.

- anonymous

So, we get|dw:1338103103383:dw|

- anonymous

A) answer is 12

- anonymous

More specifically, \(t=\pm 2\).

- anonymous

a)=-12m/s

- anonymous

But, "t" cannot be negative, so t = 2.
so, c) 2

- anonymous

A) -12m/s correct. sorry for my error.

- anonymous

On double differentiation, we get, a = 6t (I think I am correct).

- anonymous

So, unifrom acceleration

- anonymous

@Ishaan94 , @apoorvk @him1618 please help me and correct me if I am wqrong.

- anonymous

Acceleration is not uniform

- anonymous

But how? My steps r wrong or rite? Please help me.

- apoorvk

No as a=6t, acceleration is dependent on time. So it's not uniform.

- anonymous

\(t\) most certainly can be negative. It simply corresponds to be fore \(t=0\).

- anonymous

Bt that isnt in the domain@ yakeyglee

- anonymous

So @apoorvk bhai, when can we say that a is uniform? in what relation?

- apoorvk

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.

- anonymous

When the accn is not a function of time

- apoorvk

uniform means constant - it isn't constant is it?

- anonymous

In this case accn is a function of time...so its not uniform

- anonymous

Thanks all, I understood.

- anonymous

But, plz show me what will be the answer of (a), (b) and (c)

- anonymous

velocity is dx/dt

- anonymous

@him1618 I don't see any statement of the domain...it's usually assumed \(t\in\mathbb R\), which includes negative values.

- anonymous

So, (a) = -12, (c) = 2, (b) = not uniform. Is this the final answer? Please say.

- anonymous

it is so my man..bt not here..this is simple/...were not into time domain circuits here :P

- anonymous

c=4

- anonymous

Is (a) and (b) what i gave the answer rite? How is (c) = 4?????????

- anonymous

v=0
means
3t^2 - 12 =0
t=2
Put t=2 in x(t)

- anonymous

Still bhaiya, we get, 20 + 2*3 - 12*2 = 2. Am I rite????????

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