<< Chapter < Page Chapter >> Page >
The motion on earth is often modified with constant acceleration due to the combination of gravity and friction forces.

Free falling bodies under gravity represents typical case of motion in one dimension with constant acceleration. A body projected vertically upwards is also a case of constant acceleration in one dimension, but with the difference that body undergoes reversal of direction as well after reaching the maximum height. Yet another set of examples of constant accelerations may include object sliding on an incline plane, motion of an aboject impeded by rough surfaces and many other motions under the influence of gravitational and frictional forces.

The defining differential equations of velocity and acceleration involve only one position variable (say x). In the case of motion under constant acceleration, the differential equation defining acceleration must evaluate to a constant value.

v = đ x đ t i


a = đ 2 x đ t 2 i = k i

where k is a positive or negative constant.

The corresponding scalar form of the defining equations of velocity and acceleration for one dimensional motion with constant acceleration are :

v = đ x đ t


a = đ 2 x đ t 2 = k

Constant acceleration

Problem : The position “x” in meter of a particle moving in one dimension is described by the equation :

t = x + 1

where “t” is in second.

  • Find the time when velocity is zero.
  • Does the velocity changes its direction?
  • Locate position of the particle in the successive seconds for first 3 seconds.
  • Find the displacement of the particle in first three seconds.
  • Find the distance of the particle in first three seconds.
  • Find the displacement of the particle when the velocity becomes zero.
  • Determine, whether the particle is under constant or variable force.

Solution : Velocity is equal to the first differential of the position with respect to time, while acceleration is equal to the second differential of the position with respect to time. The given equation, however, expresses time, t, in terms of position, x. Hence, we need to obtain expression of position as a function in time.

t = x + 1 x = t - 1

Squaring both sides, we have :

x = t 2 - 2 t + 1

This is the desired expression to work upon. Now, taking first differential w.r.t time, we have :

v = đ x đ t = đ đ t ( t 2 - 2 t + 1 ) = 2 t - 2

1: When v = 0, we have v = 2t – 2 = 0

t = 1 s

2. Velocity is expressed in terms of time as :

v = 2 t - 2

It is clear from the expression that velocity is negative for t<1 second, while positive for t>1. As such velocity changes its direction during motion.

3: Positions of the particle at successive seconds for first three seconds are :

t = 0 ; x = t 2 - 2 t + 1 = 0 - 0 + 1 = 1 m t = 1 ; x = t 2 - 2 t + 1 = 1 - 2 + 1 = 0 m t = 2 ; x = t 2 - 2 t + 1 = 4 - 4 + 1 = 1 m t = 3 ; x = t 2 - 2 t + 1 = 9 - 6 + 1 = 4 m

Graphical representation of position

4: Positions of the particle at t = 0 and t = 3 s are 1 m and 4 m from the origin.

Hence, displacement in first three seconds is 4 – 1 = 3 m

5: The particle moves from the start position, x = 1 m, in the negative direction for 1 second. At t = 1, the particle comes to rest. For the time interval from 1 to 3 seconds, the particle moves in the positive direction.

Questions & Answers

how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
what is energy?
James Reply
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
Donsmart Reply
Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
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
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

Get the best Physics for k-12 course in your pocket!

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?