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Equation: $v_f^2 = {v_0}^2 + 2a⋅d$

$d = [{(v_0 + v_f)}/2]⋅t $

There are a number of kinematic equations describing the motion of an object, where d = displacement, $d_0$ = original position, vf = final velocity, $v_0$ = initial velocity, a = acceleration, t = time


$$Δd = d_f - d_0$$

where $d_f$ is the final displacement, and $d_0$ is the initial displacement.

Where there is constant velocity, the displacement is:

$$d = d_0 + vt$$

With constant acceleration: general formula for displacement with a starting position $d_0$, initial velocity $v_0$, and acceleration a, over time t:

$$d = d_0 + v_0t + 1/2at^2$$

Where there is constant acceleration resulting in a change in velocity from $v_0$ to $v_f$, the displacement can be calculated from the average of the two velocities:

$$d = [{(v_0 + v_f)}/2].t$$

Example: a car is travelling at 20 m/s when it passes one traffic light, then accelerates for 5 seconds to reach 40m/s. How far has it travelled?

Solution: $d = [{(v_0 + v_f)}/2]⋅t = [{(20 + 40)}/2] ⋅ 5 $
$= 30 ⋅ 5 = 150m$

Graph of displacement-time

Displacement graph

The slope of the displacement-time graph is the velocity.

Speed and Velocity

Speed |v| is the magnitude of the velocity vector.

Taking the total distance divided by the total time does not give us the average speed, unless the object was always travelling in the same direction. For speed, it is more usual to refer to the 'instantaneous speed', which is a measure of how fast an object is travelling at a specific moment.

$$|v|_{inst} = {δd}/{{δt}$$

where δd is a very short distance and δt is a very short time interval.

Velocity $v↖{→}$ is a vector, and is the rate of change in displacement in a specified direction.

$$v↖{→} = {Δd}/{Δt}$$

With acceleration a over time t, the final velocity $v_f$ is:

$$v_f = v_0 + a⋅t$$

Graph of velocity-time

Velocity graph

The area under the velocity-time graph is the change in displacement.

The slope of the velocity-time graph is the acceleration.


In the example given in the graph:

$a = {Δv}/{Δt} = {1.5}/{4.5} = 0.333$ $m/{s^2}$

The area under the graph is:

$A_{rectangle} + A_{triangle} $
$= 2.5 ⋅ 4.5 + 1/2 ⋅ 1.5 ⋅ 4.5 = 14.6$ m

Calculating the displacement algebraically:

$Δd = d_0 + v_0t + 1/2at^2$
$= 0 + 2.5 ⋅ 4.5 + 1/2 ⋅ 1.5 ⋅ 4.5 = 14.6$ m


When acceleration is constant, such as for gravity in the absence of air resistance, the instantaneous acceleration is equal to the constant acceleration:

$$a = {Δv}/{Δt}$$

$a = {v_f - v_0}/{t - 0}$, so,

$$v_f = v_0 + at$$
$d = d_0 + v_0t + 1/2at^2$
$d = {v + v_0}/2t$ 
Since $t = {v - v_0}/a$, then
$d = d_0 + v_0{v - v_0}/a + 1/2a{(v - v_0)^2}/{a^2}$
$2a(d - d_0) = 2v_0v - 2{v_0}^2 + v_f^2 + 2{v_0}^2 - 2vv_0$
∴ $v_f^2 = {v_0}^2 + 2a(d - d_0)$
$$v_f^2 = {v_0}^2 + 2a⋅d$$ 

Example: a car is travelling at 20 m/s when it passes a traffic light, then accelerates at 4m/s2 till it has travelled 150m. What is its velocity at this point?

Solution: $v_f^2 = {v_0}^2 + 2a⋅d$

= $20^2 + 2⋅4⋅150 = 400 + 1200 = 1600$

$v_f = 40 m/s$

Frames of Reference

Take the situation of a girl bouncing a ball on a train. She sees the ball fall vertically. However, what does the ball's flight look like to an observer standing on a station platform as the train goes by?

It would seem to curve as it moves down. This is because the train is moving with a constant horizontal velocity, while the ball accelerates down due to gravity.

In fact, the path the ball describes through the air for the outside observer is the same as the path of a projectile fired with horizontal velocity: a parabola.

Cannon on cliff
Constant horizontal velocity and constant vertical acceleration produce a curve in the shape of a parabola.

This begs the question of what is meant by 'at rest'? Only for an observer in the same frame of reference (in our example, on the train) can an object appear at rest.

If the observer is in a different frame of reference (that is, with relative motion) to the object being observed, then the difference in velocities has to be added to any calculation of observed motion.

$$v↖{→}_{PQ} = v↖{→}_{P} - v↖{→}_{Q}$$

where $v↖{→}_{PQ}$ is the relative velocity of P with respect to Q.

Content © Andrew Bone. All rights reserved. Created : December 18, 2013 Last updated :October 2, 2015

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