Question

The position of a particle moving along $x$ -axis is related to time

t

as follow:

x = 2 t^{2} - t^{3}

, where

x

is in meters and

t

is in seconds.
(a) What is the maximum positive displacement of the particle along the

x

axis and at what instant does it attain it?
(b) Describe the motion of the particle.
(c) What is the distance covered in the first three seconds?
(d) What is its displacement in the first four seconds?
(e) What is the particle's average speed and average velocity in the first

3

seconds?
(f) What is the particles instantaneous acceleration at the instant of its maximum positive

x

displacement?
(g) What is the average acceleration between the interval

t = 2 s

to

t = 4 s ?

Accepted Answer

Given,
$x = 2 t^{2} - t^{3}$

The derivative of $x$ w.r.t. time will give instantaneous velocity.

$v = \frac{d x}{d t} = \frac{d}{d t} (2 t^{2} - t^{3}) v = 4 t - 3 t^{2}$ ......(i)

For $v = 0$
$\Rightarrow 4 t - 3 t^{2} = 0 \Rightarrow t (4 - 3 t) = 0$

$\Rightarrow t = 0$ and $t = \frac{4}{3} s$

The double derivative of $x$ w.r.t. time will give instantaneous acceleration

$a = \frac{d^{2} x}{d t^{2}} = \frac{d^{2}}{d t^{2}} (2 t^{2} - t^{3}) a = 4 - 6 t$ .......(ii)

(a) For maximum displacement, Using Maxima-Minima
$\frac{d x}{d t} = 0$
$\Rightarrow 4 t - 3 t^{2} = 0 \Rightarrow t = 0 s or t = \frac{4}{3} s$
For maxima, double derivative of Displacement function should be negative for values of time at which $\frac{d x}{d t} = 0$ , i.e., $\frac{d^{2} x}{d t^{2}} < 0$ .

$\Rightarrow 4 - 6 t = 0$

At $t = 0 : \frac{d^{2} x}{d t^{2}} = 4 - 6 (0) = + 4 (Minima)$
and at $t = \frac{4}{3} s : \frac{d^{2} x}{d t^{2}} = 4 - 6 (\frac{4}{3}) = 4 - 8 = - 4 (Maxima)$

Therefore, the particle is at maximum displacement at $t = \frac{4}{3} s$ and the corresponding displacement is

$x_{\max} = 2 {(\frac{4}{3})}^{2} - {(\frac{4}{3})}^{3} = \frac{32}{27} m$

The maximum particle displacement of the particle is $x_{\max} = \frac{32}{27} m$ and it will occur at $t = \frac{4}{3} s .$

(b) Velocity is positive when $v > 0$ or $4 t - 3 t^{2} > 0$ or $t (4 - 3 t) > 0$ i.e., $t > 0$ and $t < \frac{4}{3} s$ For acceleration to be positive, $a > 0 \Rightarrow 4 - 6 t > 0$ or $t < \frac{2}{3} s$ From $t = 0$ to $t = \frac{2}{3} s$ , the velocity $(\vec{v})$ and $(\vec{a})$ both are in the same direction. Hence, velocity of the particle increases continuously during the time interval $t = 0$ to $t = \frac{2}{3} s$ .

From $t = \frac{2}{3} s$ onward, the acceleration $(\vec{a})$ acts in negative $x$ -direction. Hence, the velocity of the particle decreases
and becomes zero at $t = \frac{4}{3} s$ . Then after, the particle moves in negative $x$ -direction.

(c) The velocity is positive between the instants $0$ and $4 / 3 s$ and is negative from $t = \frac{4}{3} s$ to the remaining $3 s$ .
Total distance $= ∣$ Displacement from $0$ to $\frac{4}{3} s | + |$ Displacement from $\frac{4}{3}$ sto $3 s ∣$

i.e.. $|x (\frac{4}{3}) - x (0)| + |x (3) - x (\frac{4}{3})|$
$= |\{2 {(\frac{4}{3})}^{2} - {(\frac{4}{3})}^{3}\} - 0| + |\{2 (3)^{2} - (3)^{3}\} - \{2 {(\frac{4}{3})}^{2} - {(\frac{4}{3})}^{3}\}|$
$= |\frac{16}{9} (2 - \frac{4}{3})| + |\{9 (2 - 3)\} - \{\frac{16}{9} (2 - \frac{4}{3})\}| = \frac{307}{27} m$

(d) Displacement between $t = 0$ to $t = 4 s$
$Δ x = x_{t = 4} - x_{t = 0} = [\{2 (4)^{2} - 4^{3}\} - 0] = - 32 m$

(e)Average speed in first $3 s$ $= \frac{Total distance travelled in first 3 sec}{Total time} = \frac{\frac{307}{27}}{3} = \frac{307}{81} m s^{- 1}$

Average velocity for first $3 s$ $= \frac{Total displacement in first 3 s}{Total time}$ $\Rightarrow \frac{x (3) - x (0)}{3 - 0} = \frac{\{2 (3)^{2} - (3^{3})\} - 0}{3}$
${(v_{a v g})}_{t = 0 to t = 3 s} = - 3 m s^{- 1}$

(f) The particle is at its maximum displacement at $t = \frac{4}{3} s$ .
$∴$ Acceleration at $t = \frac{4}{3} s (a_{t = \frac{4}{3} s}) = \{4 - 6 (\frac{4}{3})\} = - 4 m s^{- 2}$

(g) Average acceleration
$= \frac{Total change in velocity in given time interval}{Total time duration} = \frac{v (4) - v (2)}{4 - 2}$
$= \frac{[4 (4) - 3 (4)^{2}] - [4 (2) - 3 (2)^{2}]}{2} = - 14 m s^{- 2}$

B M Sharma Solutions for Chapter: Kinematics I, Exercise 17: ILLUSTRATION 4.17

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 17: ILLUSTRATION 4.17

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 17: ILLUSTRATION 4.17 with Hints & Solutions

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