Question

The displacement of a particle is given by y = a + bt + ct² - dt³. The initial velocity and acceleration are respectively

• A. b, -4d
• B. -b, 2c
• C. b, 2c
• D. 2c, -4d

Answer 1

Answer: (C)

b, 2c

Answer 2

The displacement is given by $y = a + bt + ct^2 - dt^3$.

1. Velocity: Differentiate displacement with respect to time: $v = \frac{dy}{dt} = \frac{d}{dt}(a + bt + ct^2 - dt^3) = b + 2ct - 3dt^2$.

2. Initial Velocity: Substitute $t=0$ into the velocity equation: $v_{initial} = b + 2c(0) - 3d(0)^2 = b$.

3. Acceleration: Differentiate velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(b + 2ct - 3dt^2) = 2c - 6dt$.

4. Initial Acceleration: Substitute $t=0$ into the acceleration equation: $a_{initial} = 2c - 6d(0) = 2c$.

Therefore, the initial velocity is $b$ and the initial acceleration is $2c$.