Question

The position of a point in time $'t'$ is given by $x = a + bt - c{t^2}$, $ y = at + b{t^2}$ its acceleration at the time $'t'$ is
(A) $b - c$
(B) $b + c$
(C) $2b - 2c$
(D) $2\sqrt {{b^2} + {c^2}} $

Answer 1

Hint: Here given that position of the point at a time $'t'$ by two different coordinates as the acceleration is second-order derivatives of position concerning time. Hence differentiating the given coordinate two times will give the acceleration and after finding the acceleration for each coordinate we will deduce the resultant acceleration.

Formula used
$ \Rightarrow \vec a = \dfrac{{{d^2}\vec x}}{{{d^2}t}}$
\[ \Rightarrow a = \sqrt {a_x^2 + a_y^2} \]

Complete step by step solution:
We know that acceleration means the rate of change of velocity of an object concerning time. Its SI unit is $\dfrac{m}{{{s^2}}}$.
Given that the position of a point in x and y coordinates separately in time $'t'$ such that
$ \Rightarrow x = a + bt - c{t^2}$
$ \Rightarrow y = at + b{t^2}$
As we know that the acceleration is a second-order derivative of position for time. So acceleration can be given by the formula
$ \Rightarrow \vec a = \dfrac{{d\vec v}}{{dt}}$
$ \Rightarrow \vec a = \dfrac{{{d^2}\vec x}}{{{d^2}t}}$
Therefore, differentiating x- coordinate concerning time two times,
 $ \Rightarrow {\vec a_x} = \dfrac{{{d^2}\vec x}}{{{d^2}t}}$
$ \Rightarrow {\vec a_x} = \dfrac{{{d^2}(a + bt - c{t^2})}}{{{d^2}t}}$
$ \Rightarrow {\vec a_x} = \dfrac{{d(b - 2ct)}}{{dt}}$
Differentiating again concerning the time we will get the acceleration due to x-coordinate

$\therefore {\vec a_x} = - 2c$---------- Equation (1)
Similarly, differentiating y- coordinate concerning time two times we will get the acceleration due to y-coordinate
$ \Rightarrow {\vec a_y} = \dfrac{{{d^2}\vec y}}{{{d^2}t}}$
$ \Rightarrow {\vec a_y} = \dfrac{{{d^2}(at + b{t^2})}}{{{d^2}t}}$
$ \Rightarrow {\vec a_y} = \dfrac{{d(a + 2bt)}}{{dt}}$
Differentiating again concerning the time
$\therefore {\vec a_y} = 2b$---------- Equation (2)
Now we can find the resultant acceleration as
$ \Rightarrow {a^2} = a_x^2 + a_y^2$
\[ \Rightarrow a = \sqrt {a_x^2 + a_y^2} \]---------- Equation (3)
Now substituting the values of Equation (1) and Equation (2) in Equation (3), we get
\[ \Rightarrow a = \sqrt {a_x^2 + a_y^2} \]
\[ \Rightarrow a = \sqrt {{{( - 2c)}^2} + {{(2b)}^2}} \]
\[ \Rightarrow a = \sqrt {4{c^2} + 4{b^2}} \]
\[ \Rightarrow a = 2\sqrt {{c^2} + {b^2}} \]
\[\therefore a = 2\sqrt {{b^2} + {c^2}} \]
So the resultant acceleration is \[2\sqrt {{b^2} + {c^2}} \] in time$'t'$.

Hence, the option (D) is the correct answer.

Note: We can also obtain the resultant velocity for the given values of coordinates. As velocity is a first-order derivative of the position of an object concerning time. Here we have noted that velocity, as well as acceleration, are vector quantities so they require magnitude as well as direction also.