<< Chapter < Page Chapter >> Page >

Distance in the interval t = 0 to 1 s is :

s 1 = 1 - 0 = 1 m s 2 = 4 - 0 = 4 m

Total distance is 1 + 4 = 5 m.

6: Velocity is zero, when t = 1 s. In this period, displacement is 1 m.

7: In order to determine the nature of force on the particle, we first determine the acceleration as :

a = đ v đ t = đ đ t ( 2 t - 2 ) = 2 m / s 2

Acceleration of the motion is constant, independent of time. Hence, force on the particle is also constant during the motion.

Equation of motion for one dimensional motion with constant acceleration

The equation of motions in one dimension for constant acceleration is obtained from the equations of motion established for the general case i.e. for the three dimensional motion. In one dimension, the equation of motion is simplified ( r is replaced by x or y or z with corresponding unit vector). The three basic equations of motions are (say in x - direction) :

1: v = u + a t 2: v avg = ( u + v ) 2 3: Δ x i = u t + 1 2 a t 2

Significantly, we can treat vector equivalently as scalars with appropriate sign. Following is the construct used for this purpose :

    Sign convention

  • Assign an axis along the motion. Treat direction of axis as positive.
  • Assign the origin with the start of motion or start of observation. It is, however, a matter of convenience and is not a requirement of the construct.
  • Use all quantities describing motion in the direction of axis as positive.
  • Use all quantities describing motion in the opposite direction of axis as negative.

Once, we follow the rules as above, we can treat equations of motion as scalar equations as :

1: v = u + a t 2: v avg = ( u + v ) 2 3: Δ x = u t + 1 2 a t 2

Constant acceleration

Problem : A car moving with constant acceleration covers two successive kilometers in 20 s and 30 s respectively. Find the acceleration of the car.

Solution : Let "u" and "a" be the initial velocity and acceleration of the car. Applying third equation of motion for first kilometer, we have :

1000 = u x 20 + 1 2 a 20 2 = 20 u + 200 a 100 = 2 u + 20 a

At the end of second kilometer, total displacement is 2 kilometer (=2000 m) and total time is 20 + 30 = 50 s. Again applying third equation of motion,

2000 = u x 50 + 1 2 a 50 2 = 50 u + 2500 a 200 = 5 u + 125 a

Solving two equations,

a = - 2.86 m / s 2

Note that acceleration is negative to the positive direction (direction of velocity) and as such it is termed “deceleration”. This interpretation is valid as we observe that the car covers second kilometer in longer time that for the first kilometer, which means that the car has slowed down.

It is important to emphasize here that mere negative value of acceleration does not mean it to be deceleration. The deciding criterion for deceleration is that acceleration should be opposite to the direction of velocity.

Got questions? Get instant answers now!

Motion under gravity

We have observed that when a feather and an iron ball are released from a height, they reach earth surface with different velocity and at different times. These objects are under the action of different forces like gravity, friction, wind and buoyancy force. In case forces other than gravity are absent like in vacuum, the bodies are only acted by the gravitational pull towards earth. In such situation, acceleration due to gravity, denoted by g, is the only acceleration.

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?