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Solving problems is an essential part of the understanding

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.

Hints on solving problems

  1. Though acceleration is constant and hence one – dimensional, but the resulting motion can be one, two or three dimensional – depending on the directional relation between velocity and acceleration.
  2. Identify : what is given and what is required. Establish relative order between given and required attribute.
  3. Use differentiation method to get a higher order attribute in the following order : displacement (position vector) → velocity → acceleration.
  4. Use integration method to get a lower order attribute in the following order : acceleration → velocity → displacement (position vector).

Representative problems and their solutions

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

  • Average velocity
  • Differentiation and Integration method
  • Components of constant acceleration
  • Rectilinear motion with constant acceleration
  • Equations of motion

Average velocity

Problem : A particle moves with an initial velocity “ u ” and a constant acceleration “ a ”. What is average velocity in the first “t” seconds?

Solution : The particle is moving with constant acceleration. Since directional relation between velocity and acceleration is not known, the motion can have any dimension. For this reason, we shall be using vector form of equation of motion. Now, the average velocity is given by :

v a = Δ r Δ t

The displacement for motion with constant acceleration is given as :

Δ r = u t + 1 2 a t 2

Thus, average velocity is :

v a = Δ r Δ t = u t + 1 2 a t 2 t = u + 1 2 a t

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Differentiation and integration methods

Problem : A particle is moving with a velocity 2 i + 2 t j in m/s. Find (i) acceleration and (ii) displacement at t = 1 s.

Solution : Since velocity is given as a function in “t”, we can find acceleration by differentiating the function with respect to time.

a = đ đ t ( 2 i + 2 t j ) = 2 j

Thus, acceleration is constant and is directed in y-direction. However, as velocity and acceleration vectors are not along the same direction, the motion is in two dimensions. Since acceleration is constant, we can employ equation of motion for constant acceleration in vector form,

Δ r = u t + 1 2 a t 2 Δ r = ( 2 i + 2 t j ) t + 1 2 x 2 j x t 2

For t = 1 s

Δ r = ( 2 i + 2 x 1 j ) x 1 + 1 2 x 2 j x 1 2 Δ r = 2 i + 3 j

Note 1 : We should remind ourselves that we obtained displacement using equation of motion for constant acceleration. Had the acceleration been variable, then we would have used integration method to find displacement.

Note 2 : A constant acceleration means that neither its magnitude or direction is changing. Therefore, we may be tempted to think that a constant acceleration is associated with one dimensional motion. As we see in the example, this is not the case. A constant acceleration can be associated with two or three dimensional motion as well. It is the relative directions of acceleration with velocity that determines the dimension of motion – not the dimension of acceleration itself.

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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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