Where on the Graph Is the Velocity V>0?

Representing motion

Motion with changeless velocity

Let U.S. assume that a pushcart is moving with constant hotfoot of 2 m/s in the positive x-direction and that at t = 0 it passes through the origin.
We can stage this motion in various ways.
We May use a chemical formula and write out

v = ∆x/∆t,  x_f - x_i = v*(t_f - t_i).

If we choose our reference system so the cart is at position x = 0 at time t = 0, so x(t) = v*t.
We say that the lieu x increases linearly with the clock t,  x = (2 m/s)*t.
We sack construct the put of below.

Time (t)	Position (x)
1 s	2 m
2 s	4 m
3 s	6 m
4 s	8 m
5 s	10 m

We may also represent the motion exploitation a position versus time graph or a velocity versus sentence graph.  A pose versus prison term for our cart is shown below.
The instantaneous velocity v(t) = ∆x/∆t as ∆t becomes infinitesimally small is adequate the slope of the position versus clock graph at time t.

For motility with consistent velocity in one and only dimension the military position versus time graph is a straight line.  The slope ∆x/∆t of this straight line is equal to v.  The velocity versus time graph yields a aboveboard line with zero slope.  A velocity versus time graphical record for our cart is shown below.

Problem:

At t = 1 s, a mote aflare with constant velocity is located at x = -3 m, and at t = 6 s the particle is located at x = 5 m.
(a) From this entropy, plot the position as a social function of time.
(b) Fix the velocity of the particle from the slope of this graph.

Solution:

• Reasoning:
  For motion with uniform velocity in uncomparable dimension the position versus prison term graph is a straight line.  The slope ∆x/∆t of this straight line is equal to v.
• Details:
  (a) When the velocity is constant, the instantaneous velocity is equal to the moderate speed.
  
  (b) The average velocity of the particle is
  v = (x_f - x_i)/(t_f - t_i) = (5 m - (-3m))/(6s - 1 s) = (8/5) m/s.

---

Motion with not-uniform velocity

When an object moving in one dimension is accelerating, past

• the position versus time graph is NOT a straight line,
• the velocity versus metre graph is NOT a straight blood line with zero slope.

Links:
Positioning versus time graphs
Velocity versus time graphs

---

Motion with perpetual acceleration in one dimension

Rent out USA presume at t = 0 a cart is leaving the beginning with aught initial velocity and constant acceleration of 2 m/s² in the irrefutable x-direction.  Then

a = ∆v/∆t,  v_f - v_i = a*(t_f - t_i).

If the cart has velocity v = 0 at metre t=0, so

v(t) = a*t.

We may say that v increases linearly with the time t, v= (2 m/s²)t.
We can conception the table below.

t	vx
1 s	2 m/s
2 s	4 m/s
3 s	6m/s
4s	8 m/s
5 s	10 m/s

A velocity versus time graph for the cart is shown below.
The instantaneous acceleration a(t) = ∆v/∆t as ∆t becomes infinitesimally small is equal to the gradient of the velocity versus time graph at meter t.

For motion with constant speedup in one dimension the velocity versus clip chart is a straight line.  The slope of this straight line yields a.
The acceleration versus time graph yields a straight stemma with zero slope.

Note:  When representing motion using a graph, make a point you label the axes properly.  When presented with a gesture graph first look at the axes to identify what is premeditated.

---

Summary:

We can correspond one-dimensional motility victimisation a position versus clock time graph or a velocity versus time graph.

Position versus time chart:
The slope gives the instant velocity. (v_x = lim_∆t-->0∆x/∆t )

Positivist slope --> positive velocity
Disinclined slope --> negative velocity
Perpetual gradient --> constant velocity
Changing slope --> acceleration
(The position versus time graphical record of motion with uninterrupted speedup is a section of a parabola.)

Velocity versus metre graph:
The slope gives the instantaneous acceleration. (a_x = lim_∆t-->0∆v_x/∆t )
Sensationalism slope --> positive speedup
Negative slope --> negative speedup
Constant slope --> constant acceleration

Where on the Graph Is the Velocity V>0?

Source: http://labman.phys.utk.edu/phys221core/modules/m1/representing_motion.html

Share this post