Question

A body is moving according to the equation x = at + $b{{t}^{2}}$- c${{t}^{3}}$ where x = displacement and a, b and c are constants. The acceleration of the body is
(a) a + 2bt
(b) 2b + 6ct
(c) 2b – 6ct
(d) 3b - 6$c{{t}^{2}}$

Answer 1

In this question, an equation is given of a body. The equation is x = at + $b{{t}^{2}}$- c${{t}^{3}}$. Here, x is the displacement and ‘a’, ‘b’ and ‘c’ are the constants. We have to find the acceleration of the body. To know the acceleration, we have to find the instantaneous velocity of the body. Acceleration of any object is the rate of change in velocity at a particular time.

Complete step by step solution:
Now, let’s discuss the question.
As we all know about the velocity. It is nothing but speed of an object in a particular direction. When we say instantaneous velocity, it means velocity of an object in motion at a specific point in time. It is also said that the rate of change of displacement i.e. $\dfrac{dx}{dt}$. ‘x’ is the displacement and it is changing with respect to a time. Similarly if we talk about acceleration, it is the rate of change in velocity at a particular time i.e. $\dfrac{dv}{dt}$. ‘v’ is the velocity and it is changing with respect to a time. Now, let’s assume that for a short interval of time,
$\Rightarrow$v = $\dfrac{dx}{dt}$
So,
$\Rightarrow \dfrac{dv}{dt}$ = $\dfrac{d\left( \dfrac{dx}{dt} \right)}{dt}$
On solving:
$\Rightarrow \dfrac{d\left( \dfrac{dx}{dt} \right)}{dt}=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$
Now, $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$ is nothing but acceleration.
Let’s write the given equation of the body now.
$\Rightarrow$x = at + $b{{t}^{2}}$- c${{t}^{3}}$
Differentiate the equation with respect to ‘t’, we get:
$\Rightarrow \dfrac{dx}{dt}=a+2bt-3c{{t}^{2}}$
Again differentiate with respect to ‘t’ to find the acceleration as we know that $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$ is the acceleration.
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=2b-6ct$

So, the correct answer is “Option c”.

Note: Students make sure that we are using differentiation rules in above answer. We are using: Power rule: $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
Derivative of a constant: $\dfrac{d\left( a \right)}{dx}=0$
While doing double derivative, we have to solve for the result of single derivative. We don’t have to again repeat the derivative for the given equation.