Question

A particle moves in One dimension the velocity is given by v(t) =c2t^2 +c1t +c0 , where c0, c1 , and c2 are constants . The acceleration of the particle at t = 1 second isplease answer it ​

Answer 1

GiveN :

• Velocity of a particle is given as $\sf{v(t) \: = \: c_2 t^2 \: + \: c_1 t \: + \: c_0 \: ms^{-1}}$
• And $\sf{c_0 \: , \: c_1 \: and \: c_2}$ are constant.

To FinD :

• Acceleration at t = 1 s

SolutioN :

Take the given equation :

$\longrightarrow \sf{v \: = \: c_2 t^2 \: + \: c_1 t \: + \: c_0} \\ \\ \small{\sf{\: \: \: \: \: \: \: \: Differentiate \: wrt. \: t}} \\ \\ \longrightarrow \sf{\dfrac{dv}{dt} \: = \: \dfrac{d(c_2 t^2 \: + \: c_1 t \: + \: c_0)}{dt}} \\ \\ \longrightarrow \sf{a \: = \: 2 c_2 t \: + \: c_1 \: + \: 0} \\ \\ \longrightarrow \sf{a \: = \: (2 c_2 t \: + \: c_1) \: ms^{-2}}$

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Now, put t = 1s

$\longrightarrow \sf{a \: = \: 2 c_2 (1) \: + \: c_1} \\ \\ \longrightarrow \sf{a \: = \: (2c_2 \: + \: c_1) ms^{-2}} \\ \\ \underline {\sf{\therefore \: Acceleration \: at \: t \: = \: 1 \: s \: is \: (2c_2 \: + \: c_1) ms^{-2}}}$

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Additional Information :

• Acceleration is defined as rate of change of velocity.

• Acceleration of denoted by a.

• SI unit of acceleration is m/s².

• Acceleration = Final velocity - Initial velocity/Time

• Instantaneous acceleration is given by dv/dt

Answer 2

Given :-

Velocity of particle moving in one dimension =

$v(t) = c2 {t}^{2} + c1t + c0$

And C1 , C2 , C0 are constant

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Find :-

Acceleration of particle at time t = 1 second

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Solution :-

Acceleration of particle = dv/dt

It means for finding acceleration we have to find differentiation of given velocity

$\frac{dv}{dt} = \frac{d(c2 {t}^{2} + c1t + c0)}{dt}$

$acceleration = 2 \times c2t + c1 + 0........(1)$

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Now we have to find Velocity at time , t = 1 second

Put value of t in equation

{$2 \times c2(1) + c1$} m/s^2

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So here is Your answer

Acceleration of moving particle in one dimension at time t = 1 second is

{$2 \times c2(1) + c1$} m/s^2