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Rectilinear Motion by Albin Gomes - msdelacruzinfo

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									Name- Albin Gomes
Ms. Dela Cruz
Introduction:
A motion along straight line is called rectilinear motion. In general, it need not be one –
dimensional; it can take place in a two dimensional plane or in three dimensional space. But, it
is always possible that rectilinear motion be treated as one dimensional motion, by suitably
orienting axes of the coordinate system. This fact is illustrated here for motion along an
inclined plane. The figure below depicts a rectilinear motion of the block as it slides down the
incline. In this particular case, the description of motion in the coordinate system, as shown,
involves two coordinates (x and y).




The speed and velocity in rectilinear motion are given as:

V(t) = s’(t) = ds/dt


S(t) = / s’(t) / = / ds/dt /


Acceleration:

A(t) = v’(t) = dv/dt

A(t) = s”(t) = d^s/dt^2
Speeding up and slowing down:

A particle in rectilinear motion is speeding up when its speed is increasing and slowing down
when the speed is decreasing.
    When both, the velocity and acceleration have positive signs; the particle is speeding up.
    When the velocity and acceleration have opposite signs, the particle is slowing down.
    When velocity equals zero and acceleration is not equal to zero, there is a change in
      direction.


Analyzing position versus time curve:

The position versus time curve contains the following information:

    If s(t) is positive, the particle is on the positive side of s-axis.
    If s(t) is negative, the particle is on the negative side of s-axis.
    The slope of this curve at any instant represents the instantaneous velocity at that time.
    When curve has a positive slope, velocity is positive and the particle is moving in an
     opposite direction.
    When the curve has a negative slope, the velocity is negative and the particle is moving
     towards negative direction.
    When the curve is zero, the particle has momentarily stopped.

 Important definitions:

        Velocity is defined as the rate of change of position. It is a vector physical quantity, both
         speed and direction are required to define it.
        Acceleration is defined as the rate of change of velocity, or, equivalently, as the second
         derivative of position.
        Speed is the rate of motion, or equivalently the rate of change in position, often expressed
         as distance d traveled per unit of time t.
        Displacement is the position of a point in reference to an origin or to a previous position.



          Example 1: Rectilinear motion
Problem: If the position of a particle along x – axis varies in time as :
x=2t2−3t+1
Then:

   1. What is the velocity at t = 0?
   2. When does velocity become zero?
   3. What is the velocity at the origin?

Solution: We first need to find out an expression for velocity by differentiating the given
function of position with respect to time as:

V= ds/dt (2t^2- 3t + 1) = 4t- 3

   1. the velocity at time t = 0
     4(0) – 3 = -3ft.

   2. the velocity becomes 0 at:
      4t – 3 = 0
       t = ¾ = .75sec

   3. velocity at origin:
      at origin x = 0
      2t^2 – 3t + 1 = 0
      (2t – 1) ( t – 1) = 0
      t = .5, 1 sec
      v(.5) = 4(.5) – 3 = -1
      v(1) = 4(1) – 3 = 1

Example 2: Rectilinear motion
Problem: Find the acceleration and velocity function of : s(t) = t^3 – 6t^2.

Solution:
v(t) = ds/dt(t^3 – 6t^2) = 3t^2 – 12t

a(t) = d^2s/dt^2(t^3 – 6t^2) = 6t – 12
Example 3: Rectilinear motion
Problem: an object is thrown up from a height of 16ft, with an initial velocity of 24 ft/sec.
Its height after 1 sec. is given by: 16 + 24t – 16t^2; 0 < t < 2.
   a. Find velocity at time t?
   b. When is the velocity 0?
   c. When is velocity positive?
   d. When is velocity negative?
   e. When is the object highest? How high?

Solution:
   a. v = -32t + 24.
   b. v = 0
      - 32t+ 24 = 0
       T = 24/32 = ¾ sec.
   c. Velocity is positive for time [ 0, ¾) because v(t) > 0.
   d. Velocity is negative for time (3/4, 2] because v(t) < 0.
   e. y(3/4) = 16 + 24(3/4) -16(3/4)^2
            = 16 + 18 – 9 = 25 ft. high. Highest point at ¾ secs.
1.   WIKIPEDIA.COM
2.   GOOGLE.COM
3.   CNX.ORG
4.   GOOGLE/IMAGES
5.   CALCULAS EARLY TRANSCEDENTALS BY ANTON.

								
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