The displacement of the particle is given by y=a+bt+ct²-dt⁴. The initial velocity and acceleration are respectively

To find the initial velocity and initial acceleration of the particle, we start with the given equation for displacement:
$y=a+bt+c{t}^2-d{t}^4$

Step 1: Find the velocity of the particle
Velocity ($v$) is the first derivative of displacement ($y$) with respect to time ($t$):
$v=\frac{dy}{dt}$
Differentiating each term of the displacement equation with respect to $t$:
$v=\frac{d}{dt}(a+bt+c{t}^2-d{t}^4)$
$v=0+b+2ct-4d{t}^3$
$v=b+2ct-4d{t}^3$

Step 2: Find the initial velocity
Initial velocity ($v_0$) is the velocity at time $t=0$:
$v_0=b+2c\left(0\right)-4d{\left(0\right)}^3=b$

Step 3: Find the acceleration of the particle
Acceleration ($a$) is the first derivative of velocity with respect to time (or the second derivative of displacement):
$a=\frac{dv}{dt}$
Differentiating the velocity equation with respect to $t$:
$a=\frac{d}{dt}(b+2ct-4d{t}^3)$
$a=0+2c-12d{t}^2$
$a=2c-12d{t}^2$

Step 4: Find the initial acceleration
Initial acceleration ($a_0$) is the acceleration at time $t=0$:
$a_0=2c-12d{\left(0\right)}^2=2c$

Therefore, the initial velocity is b and the initial acceleration is 2c.
The correct option is b, 2c.