Question

The velocity of a body as a function of time is $V = {t^3} - 6{t^2} + 10t + 4$. The correct increasing order of accelerations of a body at the given following time instant will be
i) t = 0 sec
ii) t =1 sec
iii) t= 5 sec

A) ii, i, iii
B) i,ii, iii
C) iii, ii, i
D) iii, i, ii

Answer 1

Hint: The accelerations of a body can be calculated by differentiating the velocity of a body as a function of time.

Formula Used:
$Acceleration(a) = \dfrac{{dV}}{{dt}}{\text{ }}$, where $V$ is velocity of body as a function of time ($t$).
Formula of Differentiation: $\dfrac{{d{x^n}}}{{dt}} = n{x^{n - 1}}$, where $x$: variable as a function of time($t$), $n$ is order of differentiation.

Complete step by step answer:
Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.
Mathematically, acceleration can be defined as: $Acceleration(a) = \dfrac{{dV}}{{dt}}{\text{ }}$, where $V$ is velocity of body as a function of time (t).
Given, the velocity of a body as a function of time is $V = {t^3} - 6{t^2} + 10t + 4$……….(i)
As, $a = \dfrac{{dV}}{{dt}}{\text{ }}$
Putting the expression of velocity from (i), we get
$ \Rightarrow a = \dfrac{{d\left( {{t^3} - 6{t^2} + 10t + 4} \right)}}{{dt}}$
Now, splitting each term, we get:
$ \Rightarrow a = \dfrac{{d\left( {{t^3}} \right)}}{{dt}} - \dfrac{{d\left( {6{t^2}} \right)}}{{dt}} + \dfrac{{d\left( {10t} \right)}}{{dt}} + \dfrac{{d\left( 4 \right)}}{{dt}}$
As we know that differentiation of the constant term is zero.
$\Rightarrow \dfrac{{d\left( 4 \right)}}{{dt}} = 0$
So,
$\Rightarrow a = \dfrac{{d\left( {{t^3}} \right)}}{{dt}} - \dfrac{{d\left( {6{t^2}} \right)}}{{dt}} + \dfrac{{d\left( {10t} \right)}}{{dt}}$
Taking out the constant term from each differentiation, we have:
$ \Rightarrow a = \dfrac{{d\left( {{t^3}} \right)}}{{dt}} - 6\dfrac{{d\left( {{t^2}} \right)}}{{dt}} + 10\dfrac{{d\left( t \right)}}{{dt}}$
Now, applying the formula of Differentiation: $\dfrac{{d{x^n}}}{{dt}} = n{x^{n - 1}}$ in above equation, we get:
\[
   \Rightarrow a = 3{t^{3 - 1}} - 6 \times 2{t^{2 - 1}} + 10{t^{1 - 1}} \\
   \Rightarrow a = 3{t^2} - 12{t^1} + 10{t^0} \\
 \]
As, we know that ${t^0} = 1$
So, \[a = 3{t^2} - 12t + 10\]……………………… (ii)

(a) t =0 sec
Now, we put the value of time (t) = 0 sec in equation (ii), we get:
\[ \Rightarrow a = 3{\left( 0 \right)^2} - 12 \times 0 + 10\]
\[ \Rightarrow a = 10m/{s^2}\]…………………….. (iii)

(b) t =1 sec
Now, we put the value of time (t) = 1 sec in equation (ii), we get:
\[
   \Rightarrow a = 3{\left( 1 \right)^2} - 12 \times 1 + 10 \\
   \Rightarrow a = 3 - 12 + 10 \\
 \]
\[\therefore a = 1m/{s^2}\]……………………….. (iv)

(c) t =5 sec
Now, we put the value of time (t) = 5 sec in equation (ii), we get
\[
   \Rightarrow a = 3{\left( 5 \right)^2} - 12 \times 5 + 10 \\
   \Rightarrow a = 3 \times 25 - 60 + 10 \\
   \Rightarrow a = 75 - 50 \\
 \]
\[\therefore a = 25m/{s^2}\]………………………………… (v)

By comparing the values of acceleration from (iii), (iv) & (v),we get
$b< a < c$

Thus, answer is option (A) b,a,c

Note:
The meaning of velocity of an object can be defined as the rate of change of the object’s position with respect to a frame of reference and time. The SI unit of it is meter per second ($ms^{-1}$). If there is a change in magnitude or the direction in the velocity of a body the body is said to be accelerating.
Both velocity and acceleration are vector quantities.