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Question

(C) The position of the particle is given by $y = a + bt + ct^2 - dt^4$.The velocity $v$ is the first derivative of position with respect to time $t$:

Vedclass · Accepted Answer

$b, 2c$

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(C) The position of the particle is given by $y = a + bt + ct^2 - dt^4$.The velocity $v$ is the first derivative of position with respect to time $t$:$v = \frac{dy}{dt} = \frac{d}{dt}(a + bt + ct^2 - dt^4) = 0 + b + 2ct - 4dt^3$.At the initial time $t = 0$,the initial velocity $v_{initial}$ is:$v_{initial} = b + 2c(0) - 4d(0)^3 = b$.The acceleration $a_{acc}$ is the derivative of velocity with respect to time $t$:$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(b + 2ct - 4dt^3) = 0 + 2c - 12dt^2$.At the initial time $t = 0$,the initial acceleration $a_{initial}$ is:$a_{initial} = 2c - 12d(0)^2 = 2c$.Thus,the initial velocity is $b$ and the initial acceleration is $2c$.

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

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