Question

The equation of motion is $s=a{{t}^{2}}+bt+c$. If $s$ after 1 sec is 20 m, velocity after 2 sec is 30 $m{s^{-1}}$ and acceleration is 10 $m{s^{-2}}$, then
(a) $2a+2b=c$
(b) $a+b=c$
(c) $a+c=b$
(d) $b+c=a$

Answer 1

Hint: We have the equation of motion which is giving us the displacement $s$ at time $t$. We will find the equation for velocity by differentiating the equation for displacement. Then we can obtain the equation for acceleration by differentiating the equation of velocity. We will substitute the values given for time, displacement, velocity, and acceleration in the respective equations to obtain relations between $a$, $b$, and $c$.

Complete step-by-step solution
The given equation of motion is $s=a{{t}^{2}}+bt+c$. Now, we will differentiate this equation to get the expression for velocity $v$ at time $t$ in the following manner,
$\dfrac{ds}{dt}=\dfrac{d}{dt}\left( a{{t}^{2}}+bt+c \right)$
We know that $v=\dfrac{ds}{dt}$, that is change in displacement with respect to time. Now, to differentiate the above equation, we will use the following result,
\[\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)+\dfrac{d}{dx}\left( g\left( x \right) \right)\]
where $f\left( x \right)$ and $g\left( x \right)$ are functions in the variable $x$. Using this result, we get
$\dfrac{ds}{dt}=\dfrac{d}{dt}\left( a{{t}^{2}} \right)+\dfrac{d}{dt}\left( bt \right)+\dfrac{d}{dt}\left( c \right)$
We know that the derivative of a constant is zero. We also know that for a polynomial function, the derivative is given as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. Using this formula, we get
$v=2at+b$
We have obtained the expression for velocity $v$ at time $t$. Now, we will differentiate this expression to obtain the acceleration, since we know that $acc=\dfrac{dv}{dt}$. We have the following,
$\dfrac{dv}{dt}=\dfrac{d}{dt}\left( 2at+b \right)$
Again, we will use the result mentioned above and the formula for differentiation of a polynomial function. Applying these two on the above expression, we get
$\begin{align}
  & acc=\dfrac{d}{dt}\left( 2at \right)+\dfrac{d}{dt}\left( b \right) \\
 & =2a
\end{align}$
Now, we are given that $s$ after 1 sec is 20 m. We will substitute $s=20$ and $t=1$ in the given equation of motion as follows,
$20=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c$
Simplifying this expression we get,
$a+b+c=20....(i)$
Next, we know that $v$ after 2 sec is 30 $m{s^{-1}}$. We will substitute $v=30$ and $t=2$ in the equation we obtained for velocity in the following way,
$30=2a\left( 2 \right)+b$
Simplifying this expression, we get
$4a+b=30....(ii)$
We also know that after 2 sec, the acceleration is 10 $m{s^{-2}}$. Therefore, we will substitute the values $acc=10$ and $t=2$ in the equation we obtained for acceleration, as follows
$10=2a$
Simplifying this expression, we obtain the value of $a$, which is $a=5$.
Substituting $a=5$ in equation $(ii)$, we get
$4\times 5+b=30$
Solving for $b$, we get
$\begin{align}
  & 20+b=30 \\
 & \therefore b=10 \\
\end{align}$
Substituting the values $a=5$ and $b=10$ in equation $(i)$, we get
$\begin{align}
  & 5+10+c=20 \\
 & \therefore c=5 \\
\end{align}$
We have $a=5,\text{ }b=10,\text{ }c=5$. It is clear that $a+c=b$. Therefore, the correct option is (c).

Note: We should be familiar with the fact that velocity is the differentiation of displacement and acceleration is the differentiation of velocity. This is useful in forming equations in which we can use the information given in the question. Generally, acceleration is denoted by $a$. But we already had a variable with the same name. Hence, to avoid confusion, we denoted the acceleration by $acc$.