Showing posts with label Horizontal Rectilinear Motion. Show all posts
Showing posts with label Horizontal Rectilinear Motion. Show all posts

Differential Calculus - Rectilinear Motion

RECTILINEAR MOTION

It is a particle's motion along a straight line path. The motion of the particle was perfectly characterized by an equation of the form s=f(t), also known as the equation of motion, where t>=0 is the time and is the displacement of the particle at any given time, measured from a selected fixed point in its path known as the reference point.

The chosen reference point is typically the initial position of the particle at time zero (t = 0). The equations below can be used to calculate the particle's velocity (v) and acceleration (a) at time t.

HORIZONTAL RECTILINEAR MOTION

This is motion of the particle is along a horizontal straight line. Since displacement, velocity and acceleration are all vector quantities, the following are the sign conventions:
  • If s > 0 , right beside the reference point the particle is located.
  • If s < 0 , left beside the reference point, the particle is located. 
  • If v > 0 , the particle is moving in an increasing direction. s (moving to the right away from the reference point) 
  • If v < 0 ,the particle is moving in an decreasing direction. s (moving to the left, towards or away from the reference point)
  • if v = 0, the particle is at rest at that particular time
  • If a > 0 , velocity is increasing
  • If a < 0 , velocity is decreasing which means the particle is decelerating 

VERTICAL RECTILINEAR MOTION

A good illustration of a vertical rectilinear motion is free-fall motion. The moving particle, which is referred to as a freely falling body, is only influenced by its weight, and air resistance is considered to be negligible, It's acceleration is due to gravity.


The first two values are rounded-off to the nearest integer for computational convenience. For s and v , the following sign convention applies:
  • If s > 0 , the particle is above the reference point
  • if s < 0, the particle is below the reference point
  • if v < 0, the particle is moving in upward direction
  • if v > 0, the particle is moving in downward direction

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Problem Solving of Rectilinear Equation

The path taken by a particle is a horizontal line as it
moves according to the law s=16t4-76t3-7t2+12t+1.
At what time will its acceleration be equal to one?
What its velocity at that time? 


Step 1: Get the Derivative of s=16t4-76t3-7t2+12t+1
to get the velocity.

v=23t3-72t2-14t+12


Step 2: Get the Derivative of v=23t3-72t2-14t+12
to get the acceleration

a=2t2-7t-14


Step 3: Equate the acceleration to one
At what time will its acceleration be equal to one?

2t2-7t-14=1
2t2-7t-15=0
(2t+3)(t-5)=0t=5


Step 4: When t=5 at v=23t3-72t2-14t+12
What its velocity at that time? 

a(5)=23(5)3-72(5)2-14(5)+12
a(5)=2503-1752-1392
a(5)=2503-157
a(5)=-2213


Final Answer: t=5, a(5)=-2213



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Solving a Rectilinear Motion

A particle moves horizontally according to the law s=13t3-12t2-12t+1. 
At what velocity is the particle moving when t=1? when will the particle 
come to rest and what is its acceleration at that time?


Step 1: Get the derivative of s=13t3-12t2-12t+1 to get the velocity
v=t2-t-12


Step 2: At what velocity is particle moving when t=1?
v=(1)2-(1)-12
v=-12


Step 3: Get the factor of the velocity equation,
When will the particle come to rest?
t2-t-12=0
(t-4)(t+3)=0
t=4


Step 4: What is its acceleration at that time?
Get the derivative of velocity equation to get the acceleration
t2-t-12
a=2t-1


when t=4
a=2(4)-1
a=7


Final Answer: v=-12, t=4, a=7



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