The displacement of a particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and acceleration a

Question

The displacement of a particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and acceleration are, respectively:

Accepted Answer

1.  **Velocity ($v$):** Instantaneous velocity is defined as the derivative of displacement with respect to time ($v = \frac{dy}{dt}$) .
    Given $y = a + bt + ct^2 - dt^4$.
    Differentiating with respect to $t$:
    $v = \frac{d}{dt}(a + bt + ct^2 - dt^4)$
    $v = 0 + b + 2ct - 4dt^3$.
2.  **Initial Velocity:** At $t = 0$, substitute $t=0$ into the velocity equation:
    $v_{initial} = b + 2c(0) - 4d(0) = b$.
3.  **Acceleration ($acc$):** Instantaneous acceleration is defined as the derivative of velocity with respect to time ($acc = \frac{dv}{dt}$) .
    Differentiating $v$ with respect to $t$:
    $acc = \frac{d}{dt}(b + 2ct - 4dt^3)$
    $acc = 0 + 2c - 12dt^2$.
4.  **Initial Acceleration:** At $t = 0$, substitute $t=0$ into the acceleration equation:
    $acc_{initial} = 2c - 12d(0) = 2c$.
5.  **Conclusion:** The initial velocity is $b$ and the initial acceleration is $2c$.

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