The relationship between displacement, velocity and acceleration is summarised in the following diagram.
Given the displacement, to find the velocity, we differentiate. To find the acceleration, we differentiate twice.
Given the acceleration, to find the velocity, we integrate. To find the displacement, we integrate twice.
Example
A particle P moves on the axis. At time t seconds, its acceleration isWhen its velocity isWhenthe displacement is 2m. Find expressions in terms of for the velocity and displacement.
When
When
Example:
The displacement of a particle isFind the velocity and acceleration in terms of t. When is the particle at rest for? What are the displacement and acceleration at this time?
When the particle is at rest,
The particle is at rest whenor
The displacement whenis
The acceleration whenis
]]>The horizontal component of acceleration is zero soat every point.
The vertical acceleration isThe negative sign indicates the acceleration is always directed downwards, and is consistent with the sign convention taking upwards as positive and downwards as negative.
At the highest point, the vertical component of velocity is zero.
For example, suppose a ball is thrown horizontally from a window 6m above the ground. It just passes over a wall a horizontal distance 3m from the wind and hits the ground 4m behind it. We may draw a diagram:
We want to find the speedand the height of the wall.
We can draw up the following table for the entire motion, up until when the ball hits the ground.
Vertical 
Horizontal 

6 
7 

0 
u 


u 

9.8 
0 



We can calculate the time taken for the ball to hit the ground by considering the vertical motion.
We put this into the table in both columns, because time is the same vertically and horizontally.
Vertical 
Horizontal 

6 
7 

0 
u 


u 

9.8 
0 


1.107 
1.107 
Now we can findby considering the horizontal motion.
The time to reach the wall can now be found by considering the horizontal motion.
Vertical 
Horizontal 


3 

0 
6.326 


6.326 

9.8 
0 



Again write this time in both vertical and horizontal columns.
Vertical 
Horizontal 


3 

0 
6.326 


6.326 

9.8 
0 

0.474 
0.474 
Work out the vertical distance fallen before passing over the wall using
Hence the height of the wall is
]]>(1)
(2)
(3)
(4)
(5)
To develop the example above, suppose the ball is dropped from a height of 10m, and when it hits the ground, penetrates to a depth of 3cm.
a)Find the speed of ball when it hits the ground
b)Find the time after the ball hits the ground to come to rest and the average deceleration.
a) s=10
u=0
v=?
a=9.8
t=?
We take the negative square root because the ball is moving down.
b) s=0.03
u=14
v=0
a=?
t=?
]]>Resolving vertically gives us the equation(1)
Taking moments about A gives
Then from (1)
It is important to note that the following assumptions have been made:
The beam is uniform so the weight acts at the midpoint.
The beam is rigid and does not bend so all the distances are horizontal and add in a natural way.
]]>We can then just read off the bearing from A to B.
If A and B are given as position vectors
andthen it is useful to find the vectorand plot it, on a compass with the centre of the compass acting as A.
The bearing will be given by
ifandare both positive
ifis negative and is positive
ifandare both negative
ifis positive andis negative
]]>We can use the equilibrium condition, that all the forces in any direction add up to zero, to write down simultaneous equations and solve them for the forces we need to find.
Example:
Resolving vertically for the 2Kg particle:
(1)
Resolving horizontally
(2)
gives
Then from
]]>When a car is pulling a trailer, we can applywhereis the resultant of all the forces on both car and trailer,is the mass of the car plus the mass of the trailer, and a is the acceleration, which, since the cable between the cable and car is always assumed inextensible, is the same for both car and trailer, to the system as a whole – In the diagrams belowis the driving force...
whereis the total mass of car plus trailer.
or each individually...
Applyingfor the trailer we obtain
(1)
Applyingfor the car we obtain
(2)
(1)+(2) gives
Notice that when we draw the force diagram for the car and trailer separately:
The tensions acting on trailer and car are equal and opposite. The tension is a force internal to the system so does not appear in the top diagram, which takes only external forces into account. That they are equal and opposite is a consequence of Newton's 3^{rd} Law.
The accelerations are the same for car and trailer, the cable is inextensible and are the same as in the first diagram.
andare the same in the first and second diagrams.
]]>As shown on the diagram above, the reaction or contact force R is always perpendicular to the contact surface of the two bodies and the friction force always acts along the contact surface between two bodies. Friction always acts to oppose motion so if the force P is reversed so is the force
If we resolve vertically for the diagram we get
Resolving horizontally gives
]]>Suppose then that one particle is thrown up from the ground with a speed of 10 m/s and simultaneously another particle is dropped from a point 8 m vertically above the first one. Applying SUVAT to each particle in the direction of motion of each particle.
Particle 1 
Particle 2 

s 
? 
s 
? 
u 
10 
u 
0 
v 
? 
v 
? 
a 
9.8 
a 
9.8 
t 
? 
t 
? 
There are also two conditions we can use. Sine the particles start moving at the same time, t is the same for both, and when the particle collide, they will have moved a total of 8 m so that when the particles collide, if particle 1 has moved a distance y, particle 2 will have moved a distance 8y. The table becomes
Particle 1 
Particle 2 

s 
y 
s 
8y 
u 
10 
u 
0 
v 
? 
v 
? 
a 
9.8 
a 
9.8 
t 
t 
t 
t 
For particle 1, usinggives
For particle 2, usinggives
Adding these two equations givesand this collision point is independent of the acceleration due to gravity.
In fact the particles are at each moment of the motion moving together with combined speed 10. If the initial combined approach speed of both particles isand the distance between them is then they will collide after a time
]]>IF THE TWO PARTICLES START AT DIFFERENT TIMES WE CANNOT USE THE SAME TIME WITHOUT MODIFICATION!
For particle B we can write
here for particle B is not the same asfor particle A because particle B started a time later than particle A. In factis the time for which particle B has been travelling and since B startedseconds later than A,
We can write thereforeand
We can use this in the following example.
A particle A starts atfromwith velocityand a particle B starts at fromwith velocityFind the distance between the particles when
then
then
The displacement vector from A to B isWhenthe displacement vector isand the distance between A and B whenis
]]>