Question

The displacement of a particle is given by $y = a + bt + c{t^2} - d{t^4}$. The initial velocity and acceleration are respectively:
(A) $b, - 4d$
(B) $ - b,2c$
(C) $b,2c$
(D) $2c, - 4d$

Answer 1

Hint: The differentiation of the displacement of a particle with respect to time gives us the velocity of that particle at that time. To find the initial velocity we need to substitute the value of time as zero. We can get the acceleration of a body by differentiating the velocity of the body with respect to the time and substituting the time to zero.
Formula used: In the solution to this question we will be using the following formulas,
$\Rightarrow v = \dfrac{{dy}}{{dt}}$
where $v$ is the velocity of the particle and $y$ gives the displacement.
and, $a = \dfrac{{dv}}{{dt}}$
where $a$ is the acceleration of the particle.

Complete step by step answer:
We are provided with the displacement of a particle as a function of time as,
$\Rightarrow y = a + bt + c{t^2} - d{t^4}$
Now the velocity can be found out from the displacement of any particle by differentiating it with respect to time, that is,
$\Rightarrow v = \dfrac{{dy}}{{dt}}$
So we put $y$ in the above equation and get,
$\Rightarrow v = \dfrac{d}{{dt}}\left( {a + bt + c{t^2} - d{t^4}} \right)$
Now differentiating every term we get,
$\Rightarrow v = b + 2ct - 4d{t^3}$
To get the value of the initial velocity we substitute the value of time as $t = 0$. Hence we get,
$\Rightarrow {\left. v \right|_{t = 0}} = b + 2c \times 0 - 4d \times 0$
so we get the initial velocity as,
$\Rightarrow {\left. v \right|_{t = 0}} = b$
The acceleration of any body is given by the rate of change of velocity, that is we need to differentiate the velocity with respect to time in order to get the acceleration. Therefore,
$\Rightarrow a = \dfrac{{dv}}{{dt}}$
So by substituting the value of the velocity we get,
$\Rightarrow a = \dfrac{d}{{dt}}\left( {b + 2ct - 4d{t^3}} \right)$
On performing differentiation on every term we get,
$\Rightarrow a = 2c - 12d{t^2}$
To get the value of the initial acceleration we substitute the value of time as, $t = 0$. Hence we get,
$\Rightarrow {\left. a \right|_{t = 0}} = 2c - 12d \times 0$
Therefore the initial acceleration is,
$\Rightarrow {\left. a \right|_{t = 0}} = 2c$
So the initial velocity and acceleration of the particle are $b$ and $2c$ respectively.
Therefore the correct option is (C).

Note:
The velocity of a body depends on the initial and the final positions of the body. So the velocity is the displacement of the body with respect to time. The velocity of the body can be zero when the initial and the final positions of a body are the same. But a body having zero velocity can still have non-zero acceleration.