# If $s=2{{t}^{3}}-6{{t}^{2}}+at+5$ is the distance travelled by a particle at time ‘t’ and if the velocity is ‘-3’ when its acceleration is zero, then the value of ‘a’ is

(a) -3

(b) 3

(c) 4

(d) -4

(e) 2

Last updated date: 18th Mar 2023

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Answer

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Hint: Differentiate the function ‘s’ with respect to ‘t’ and further differentiate it with respect to ‘t’ and then use the value of ‘a’ given to find ‘t’ and put it back to the function ‘v’ to get the value of ‘a’ as asked in the question.

“Complete step-by-step answer:”

In the question we are given the equation of ‘s’ in terms of ‘t’ where ‘a’ is constant.

$s=2{{t}^{3}}-6{{t}^{2}}+at+5........\left( i \right)$

The expression of ‘s’ in equation (i) provides information about the distance travelled by a particle in time ‘t’.

Let us first define velocity.

Velocity is the rate of distance travelled in a given time. Mathematically it is represented as

\[V=\dfrac{ds}{dt}\] which means differentiating distance with respect to time to get velocity.

Now let’s define acceleration.

Acceleration is the rate of change of velocity in a given time. Mathematically it is represented as

$a=\dfrac{dV}{dt}$ which means differentiating velocity with respect to time to get acceleration.

In equation (i) it is given that,

\[s=2{{t}^{3}}-6{{t}^{2}}+at+5\]

So by differentiating equation (i), we get velocity

We will use the formula here,

If $y={{x}^{n}}$ where ‘n’ is any rational number then,

$\dfrac{dy}{dx}=n{{x}^{n-1}}$

So we will use $V=\dfrac{ds}{dt}$,

$V=\dfrac{d}{dt}\left( 2{{t}^{3}}-6{{t}^{2}}+at+5 \right)$

$V=6{{t}^{2}}-12t+a$ ………… (ii)

So we got velocity at time ‘t’ and is equal to,

$V=6{{t}^{2}}-12t+a$

Now by further differentiating equation (ii) with respect to time we get the acceleration,

$\begin{align}

& a=\dfrac{dV}{dt}=\dfrac{d}{dt}\left( 6{{t}^{2}}-12t+a \right) \\

& a=12t-12........(iii) \\

\end{align}$

So we got acceleration at time ‘t’ and it is equal to 12t-12.

Now in the question we are given that acceleration is zero.

Now substituting a=0 in equation (iii) we get the value of t,

12t – 12 = 0

$\therefore $ t = 1

At t = 1 acceleration is ‘0’.

Now we will put t = 1s in equation (iii) we get,

$V=6{{\left( 1 \right)}^{2}}-12\left( 1 \right)+a$

$\Rightarrow $ V = 6 – 12 + a

$\Rightarrow $ V = a – 6

But in the equation we were given V = -3.

So by equating V = -3 and V = a-6 we get,

-3 = a-6

$\therefore $ a = 6-3 = 3

Hence the correct answer is option (b).

Note: Students should be careful while differentiating the functions of s, v with respect to t. They should also avoid calculation mistakes while finding and putting the values of t to get the value of a.

They should also know the formulas of differentiation by heart.

Another approach is double differentiating ‘s’, with respect to ‘t’ also gives acceleration. In this way also we will get the same answer.

“Complete step-by-step answer:”

In the question we are given the equation of ‘s’ in terms of ‘t’ where ‘a’ is constant.

$s=2{{t}^{3}}-6{{t}^{2}}+at+5........\left( i \right)$

The expression of ‘s’ in equation (i) provides information about the distance travelled by a particle in time ‘t’.

Let us first define velocity.

Velocity is the rate of distance travelled in a given time. Mathematically it is represented as

\[V=\dfrac{ds}{dt}\] which means differentiating distance with respect to time to get velocity.

Now let’s define acceleration.

Acceleration is the rate of change of velocity in a given time. Mathematically it is represented as

$a=\dfrac{dV}{dt}$ which means differentiating velocity with respect to time to get acceleration.

In equation (i) it is given that,

\[s=2{{t}^{3}}-6{{t}^{2}}+at+5\]

So by differentiating equation (i), we get velocity

We will use the formula here,

If $y={{x}^{n}}$ where ‘n’ is any rational number then,

$\dfrac{dy}{dx}=n{{x}^{n-1}}$

So we will use $V=\dfrac{ds}{dt}$,

$V=\dfrac{d}{dt}\left( 2{{t}^{3}}-6{{t}^{2}}+at+5 \right)$

$V=6{{t}^{2}}-12t+a$ ………… (ii)

So we got velocity at time ‘t’ and is equal to,

$V=6{{t}^{2}}-12t+a$

Now by further differentiating equation (ii) with respect to time we get the acceleration,

$\begin{align}

& a=\dfrac{dV}{dt}=\dfrac{d}{dt}\left( 6{{t}^{2}}-12t+a \right) \\

& a=12t-12........(iii) \\

\end{align}$

So we got acceleration at time ‘t’ and it is equal to 12t-12.

Now in the question we are given that acceleration is zero.

Now substituting a=0 in equation (iii) we get the value of t,

12t – 12 = 0

$\therefore $ t = 1

At t = 1 acceleration is ‘0’.

Now we will put t = 1s in equation (iii) we get,

$V=6{{\left( 1 \right)}^{2}}-12\left( 1 \right)+a$

$\Rightarrow $ V = 6 – 12 + a

$\Rightarrow $ V = a – 6

But in the equation we were given V = -3.

So by equating V = -3 and V = a-6 we get,

-3 = a-6

$\therefore $ a = 6-3 = 3

Hence the correct answer is option (b).

Note: Students should be careful while differentiating the functions of s, v with respect to t. They should also avoid calculation mistakes while finding and putting the values of t to get the value of a.

They should also know the formulas of differentiation by heart.

Another approach is double differentiating ‘s’, with respect to ‘t’ also gives acceleration. In this way also we will get the same answer.

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