Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$b, -4d$

$-b, 2c$

$b, 2c$

$2c, -2d$

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

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