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Q.

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

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a

$b, 2 c$

b

$b, - 4 d$

c

$- b, - 2 c$

d

$2 c, - 4 d$

answer is C.

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Detailed Solution

$y = a + bt + {ct}^{2} - {dt}^{4} ∴ v = \frac{dy}{dt} = b + 2 ct - 4 {dt}^{3} and a = \frac{dv}{dt} = 2 c - 12 {dt}^{2}$

Hence, at t = 0, v_initial= b and a_initial= 2c.

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