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Application

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the accelerated motion in two dimensions. The questions are categorized in terms of the characterizing features of the subject matter :

  • Path of motion
  • Tangential and normal accelerations
  • Nature of motion
  • Displacement in two dimensions

Path of motion

Problem : A balloon starts rising from the surface with a constant upward velocity, “ v 0 ”. The balloon gains a horizontal drift due to the wind. The horizontal drift velocity is given by “ky”, where “k” is a constant and “y” is the vertical height of the balloon from the surface. Derive an expression of path of the motion.

Solution : An inspection of the equation of drift velocity (v = ky) suggests that balloon drifts more with the gain in height. A suggestive x-y plot of the motion is shown here.

Motion of a balloon

The balloon moves with an acceleration in horizontal direction.

Let vertical and horizontal direction corresponds to “y” and “x” axes of the coordinate system. Here,

v y = v 0

v x = k y

We are required to know the relation between vertical and horizontal components of displacement from the expression of component velocities. It means that we need to know a lower order attribute from higher order attribute. Thus, we shall proceed with integration of differential equation, which defines velocity as :

đ x đ t = k y

Similarly,

đ y đ t = v 0

đ y = v 0 đ t

Combining two equations by eliminating “dt”,

x = k y đ y v 0

Now, integrating both sides, we have :

x = k y đ y v 0

Taking out constants out of the integral,

x = k v 0 y đ y

x = k y 2 2 v 0

This is the required equation of motion, which is an equation of a parabola. Thus, the suggested plot given in the beginning, as a matter of fact, was correct.

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Tangential and normal accelerations

Problem : A balloon starts rising from the surface with a constant upward velocity, “ v 0 ”. The balloon gains a horizontal drift due to the wind. The horizontal drift velocity is given by “ky”, where “k” is a constant and “y” is the vertical height of the balloon from the surface. Derive expressions for the tangential and normal accelerations of the balloon.

Solution : We can proceed to find the magnitude of total acceleration by first finding the expression of velocity. Here, velocity is given as :

v = k y i + v 0 j

Since acceleration is higher order attribute, we obtain its expression by differentiating the expression of velocity with respect to time :

a = đ v đ t = k v y i = k v 0 i

It is obvious that acceleration is one dimensional. It is evident from the data given also. The balloon moving with constant vertical velocity has no acceleration in y-direction. The speed of the balloon in x-direction, however, keeps changing with height (time) and as such total acceleration of the balloon is in x-direction. The magnitude of total acceleration is :

Questions & Answers

preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
n=a+b/T² find the linear express
Donsmart Reply
Quiklyyy
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Moment of inertia of a bar in terms of perpendicular axis theorem
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How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
Lucky Reply
work done by frictional force formula
Sudeer Reply
Torque
Misthu Reply
Why are we takingspherical surface area in case of solid sphere
Saswat Reply
In all situatuons, what can I generalize?
Cart Reply
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Clinton Reply
Explain it ?Fy=?sN?mg=0?N=mg?s
Admire Reply

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