2.7 Graphs of motion with constant acceleration (Page 6/6)

Physics for k-12 Page 6 / 6

$\begin{array}{l} a = \frac{đ v}{đ t} = - 10 m / s^{2} \end{array}$

Alternatively, we can see that expression of velocity has the form v = u + at. Evidently, the motion has a constant acceleration of -10 $m / s^{2}$ . Also, note that origin of reference and initial position are same. Applying equation of motion for position as :

$\begin{array}{l} x = u t + \frac{1}{2} a t^{2} = 40 t + \frac{1}{2} x - 10 x t^{2} = 40 t - 5 t^{2} \end{array}$

According to question, we have to find the time when particle is 60 m away from the initial position. Since it is one dimensional motion, the particle can be 60 m away either in the positive direction of the reference or opposite to it. Considering that it is 60 m away in the positive reference direction, we have :

Motion diagram

$\begin{array}{l} 60 = 40 t - 5 t^{2} \end{array}$

Re-arranging,

$\begin{array}{l} t^{2} - 8 t + 12 = 0 \\ t = 2 s or 6 s \end{array}$

We observe here that the particle is ultimately moving in the direction opposite to reference direction. As such it will again be 60 away from the initial position in the negative reference direction. For considering that the particle is 60 m away in the negative reference direction,

$\begin{array}{l} - 60 = 40 t - 5 t^{2} \end{array}$

$\begin{array}{l} \Rightarrow t^{2} - 8 t - 12 = 0 \\ t = \frac{- (- 8) \pm \sqrt {{(- 8)}^{2} - 4 x 1 x (- 12)}}{2} = - 1.29 s, 9.29 s \end{array}$

Neglecting negative value of time,

$\begin{array}{l} \Rightarrow t = 9.29 s \end{array}$

Note : We can also solve the quadratic equation for zero displacement to find the time for the particle to return to initial position. This time is found to be 8 seconds. We note here that particle takes 2 seconds to reach the linear distance of 60 m in the positive direction for the first time, whereas it takes only 9.29 – 8 = 1.29 second to reach 60 m from the initial position in the negative direction.

The particle is decelerated while moving in positive direction as velocity is positive, but acceleration is negative. On the other hand, both velocity and acceleration are negative while going away from the initial position in the negative direction and as such particle is accelerated. Therefore, the particle takes lesser time to travel same linear distance in the negative direction than in the positive direction from initial position..

Hence, the particle is 60 m away from the initial position at t = 2 s, 6 s and 9.29 s.

t (s) 0 2 4 6 8 9.29 x (m) 0 60 80 60 0 -60v (m/s) 40 20 0 -20 -40 -52.9

Motion diagram

Problem : A particle moves along x-axis with a velocity 9 m/s and acceleration -2 $m / s^{2}$ . Find the distance covered in $5^{th}$ second.

Solution : The displacement in $5^{th}$ second is :

$\begin{array}{l} x_{n} = u + \frac{a}{2} (2 n - 1) \\ x_{n} = 9 + \frac{- 2}{2} (2 x 5 - 1) \\ x_{n} = 0 \end{array}$

The displacement in the $5^{th}$ second is zero. A zero displacement, however, does not mean that distance covered is zero. We can see here that particle is decelerated at the rate of -2 $m / s^{2}$ and as such there is reversal of direction when v = 0.

$\begin{array}{l} v = u + a t \\ 2 = 9 - 2 t \\ t = 4.5 s \end{array}$

This means that particle reverses its motion at t = 4.5 s i.e in the period when we are required to find distance. The particle, here, travels in the positive direction from t = 4 s to 4.5 s and then travels in the negative direction from t = 4.5 s to 5 s. In order to find the distance in $5^{th}$ second, we need to find displacement in each of these time intervals and then sum their magnitude to find the required distance.

Motion along a straight line

The partilcle reverses its direction in 5th second .

The displacement for the period t = 0 s to 4 s is :

$\begin{array}{l} x = u t + \frac{1}{2} a t^{2} = 9 x 4 + \frac{1}{2} x 2 x 4^{2} = 36 - 16 = 20 m \end{array}$

The displacement for the period t = 0 s to 4.5 s is :

$\begin{array}{l} x = u t + \frac{1}{2} a t^{2} = 9 x 4.5 + \frac{1}{2} x 2 x {4.5}^{2} = 40.5 - 20.25 = 0.25 m \end{array}$

The displacement for the period t = 4 s to 4.5 s is :

$\begin{array}{l} {Δ x}_{1} = 20.25 - 20 = 0.25 m \end{array}$

Since particle is moving with constant acceleration in one dimension, it travels same distance on its return for the same period. It means that it travels 0.25 m in the period from t = 4.5 s to t = 5 s.

$\begin{array}{l} {Δ x}_{2} = 0.25 m \end{array}$

Thus, total distance covered between t = 4 s to t = 5s i.e. in the $5^{th}$ second is :

$\begin{array}{l} Δ x = {Δ x}_{2} + {Δ x}_{2} = 0.25 + 0.25 = 0.5 m \end{array}$

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics for k-12' conversation and receive update notifications?

Ask

	21 Kidney/Liver Biopath By Brooke Delaney Start Exam
	11 Clin Sci I - GI Twedt quiz 2 By Brooke Delaney Start Exam
	2 Neuroanatomy 02 Brain External Features By Stephen Voron Start Quiz
	Cultural Evolution Contemporary World By Richley Crapo Start Assignment
	Subject-verb Agreement By Dindin Secreto Start Quiz
	Biology Exam 2 By Vanessa Soledad Start Exam
	23 AP 23 Digestive System Essay By OpenStax Start Flashcards
	12 Dr Dowers Endocrinology By Brooke Delaney Start Exam
	Who Said This Quote Quiz By Yasser Ibrahim Start Quiz
	2011 Dynamics CRM By Danielrosenberger Start Quiz