UNIT 7. MOTION. SUMMARY
KINEMATICS
Kinematics is a branch of physics that describes the motion of objects without considering the forces that caused the motion (causes and effects).
POSITION
The position of a moving object is the place where it is located from the origin.
POSITION VECTOR
The position vector at certain time is the arrow that connects the origin with the position of the moving object. All the successive positions form the trajectory.
TRAJECTORY
All the successive positions form the trajectory. The distance travelled is the lenght covered by the object from the initial point to the final position.
DISPLACEMENT VECTOR
The displacement vector is the vector that connects the initial position with the final position in the shortest way (a straight line). It just gives information about the initial and the final position, but not about the covered path. It is the difference between the position vectors at points 1 and 2.
The displacement only has the same value than the covered space when the trajectory is a straight line.
UNIFORM LINEAR MOTION (MU) (constant speed, v)
Position-time graph
The moving object goes backwards.
The moving object goes forwards.
There is no movement. It stays still.
UNIFORMLY ACCELERATED MOTION (MUA) (constant acceleration, a) Position-time graph
It goes forwards and accelerates (speed increases, as the slope increases when time passes by)
It goes forwards and decelerates (speed decreases, as the slope decreases when time passes by)
It goes backwards and accelerates (speed increases, as the slope increases when time passes by)
It goes backwards and decelerates (speed decreases, as the slope decreases when time passes by)
Speed-time graph
Speed increases (a>0, it accelerates) Speed decreases (a<0, it brakes/decelerates)
Acceleration-time graph
a > 0 (speed increases; it accelerates)
a < 0 (speed decreases; it brakes/decelerates)
GRAPHICAL PROCEDURE for calculating distance with a speed-time graph
You can calculate the distance travelled with a speed-time graph. You just have to calculate the area under the graph.
REFERENCE FRAME OR SYSTEM OF REFERENCE
The reference frame is a coordinate system that we must choose in order to locate the moving object. When we set it, we are choosing the origin and in linear motion it is convenient to choose one of the axes along the path. In this way the vector quantities are oriented along the axis and we must assign a positive or a negative sign to indicate the direction. When the vectors have the same orientation than the axes, they are positive. When they have opposite orientation, they are negative.
HORIZONTAL MOTION. SIGN CONVENTION
Position Velocity Acceleration
VERTICAL MOTION. SIGN CONVENTION
Position Velocity Acceleration
VERTICAL MOTION
Vertical motion is always a uniformly accelerated linear motion with a value of acceleration of a=g= 9,8 m/s2.
From a certain height:
- An object is released (vi = 0 m/s)
- An object is thrown vertically downwards.
From the ground:
- An object is thrown vertically upwards. When an object is thrown vertically, the maximum height is reached when the final speed
is 0 m/s. Maximum height vf = 0
When it is said that an object is dropped, the initial speed is 0. Otherwise, they will say that the object is thrown and they would give us a value of the initial speed or we will have to calculate it. An object is dropped vi= 0
Careful, when an object falls and touches the ground do not think that the final speed is 0. If the final speed is asked, we have to assume that it refers to the speed when the object is about to touch the ground.
SAFETY DISTANCE ON THE ROAD
The safety distance is the total distance it takes a driver must keep between his/her vehicle and the vehicle in front in order to avoid collision if the car in front brakes or stops. The safety distance is the sum of:
- The reaction distance. It is the space the vehicle travels during the driver’s reaction time (time it takes the driver to react to a problem and apply the brakes). The driver has not stepped on the brake yet. We will assume the movement is a uniform motion.
- The braking distance. It is the space the vehicle travels while the driver is stepping on the brake (from the moment the brakes are first applied to the time the car comes to a complete stop), so speed decreases and there is a uniformly accelerated motion (a<0).
TYPES OF ACCELERATION
Acceleration has two components. It is a change in velocity, either in its magnitude (speed) or in its direction, or both.
(Tangential) acceleration. It is tangent to the path. It is a change in the value of velocity (it increases or decreases with time).
Normal or centripetal acceleration. It is perpendicular to the tangent to the path. It is a change in direction of velocity (it occurs when a particle moves through a curved trajectory).
UNIFORM CIRCULAR MOTION (MCU) (constant angular speed, )
1 turn= 2· radA radian is a unit of measure for angles. It is an angle whose arc length is the same as the radius. 1 radian is equivalent to 360º.
f= frequency (Hz, s1, 1/s). It is the number of turns the moving object makes in 1 second.
T= period (s). It is the time it takes an object to complete 1 turn. = angular speed (rad/s)
v= linear speed (m/s) R= radius (m)
Points A and B have the same angular velocity (both cover the same angle in the same time), but a different linear velocity v (point A moves faster, as it covers more distance in the same time than B).
The linear speed of the object is linearly proportional to the distance from the centre. When the radius from the centre increases, the linear speed increases as well. As we move to the centre, the speed decreases.
To represent the linear speed we have to draw it tangent to the path of the object.
The direction of the centripetal acceleration is toward the centre of the circumference, perpendicular to the velocity.
Example: how to change 1200 rpm to rad/s (a revolution is a turn, una revolución es una vuelta)
HOW TO SOLVE CHASE/COLLISION PROBLEMS (PROBLEMAS DE ENCUENTROS)
In these type of problems there are two moving objects (cars, motorbikes, people…) that start from different places or moments and, at certain time will meet each other (they collide or one overtakes the other one). The objective is to calculate where and when they will meet (final position and time).
Steps to solve chase/collision problems
1. Choose a reference frame (where you set the origin 0 and what orientation the X axis has) and do a drawing of the situation.
2. Write the position-time equations for both moving objects taking into account the reference frame to know their initial position and the sign of velocity (positive if it moves in the orientation of X axis or negative if it moves on the opposite orientation). If it is a uniformly accelerated motion, you must be careful with the sign of the acceleration.
It moves to the right faster (velocity increases) It moves to the right in a slower way (velocity decreases) It moves to the left in a faster way (velocity increases) It moves to the left in a slower way ( velocity decreases) When an object is braking, velocity and acceleration have opposite orientations. Remember the position-time equations for the two kind of motions we have studied:
UNIFORM MOTION (MU) UNIFORMLY ACCELERATED
MOTION (MUA)
3. Once you have the equations for each moving object, you will have a system of two equations with two variables you do not know (final position and time).
4. To solve the system, it is easier to equal the equations to each other (xf), since when they find each other their final position is the same. Then, you can get the value of time and in any of the equations you can solve the final position, which will be the same for both of them.
5. Be careful with the units (they must be consistent). For example, if the position is expressed in m, then the speed must be in m/s. If the position is expressed in km, speed must be in km/h.
CHASE/COLLISION PROBLEMS
Solved example #1. Two objects with the same starting point moving with the same orientation and starting at the same time.
Two motorbikes A and B are 10 km away. They start their movement at the same instant with constant speeds of 60 km/h and 50 km/h, respectively, with opposite orientations. Calculate when and where they will meet each other.
1. We choose our reference frame and do a drawing.
2. We write the position-time equations for each moving object. Both of them have uniform motion in this case (xf = xi + v·t)
xA = 0 + 60·t
xB = 10 50·t
3. When they find each other, xA = xB, so you can equal both equations to each other:
0 + 60·t= 10 50·t
You solve it: 10=110·t t=0,091 h
You can substitute this time in any of the two position-time equations (with one is enough). xA= 0 + 60·0,091=5,45 km xB= 1050·t=5,45 km
They will find each other 0,091 h later and at 5,45 km from where A starts.
Solved example #2. Two objects initially separated certain distance moving with the same orientation and starting at the same time.
Two motorbikes A and B are separated 5 km away. They start moving at the same time with an initial speed of 35 km/h and 25 km/h, respectively, in the same direction (from A to B). Calculate when and where they will meet each other.
1. We choose our reference frame and do a drawing.
2. We write the position-time equations for each moving object. Both of them have uniform motion in this case (xf = xi + v·t).
xA = 0 + 35·t
xB = 5 + 25·t
3. When they find each other, xA = xB, so you can equal both equations to each other: 0 + 35·t= 5 + 25·t
You solve it: 10·t=5 t=0,5 h
You can substitute this time in any of the two position-time equations (with one is enough): xA= 0 + 35·0,5=17,5 km xB= 5+25·0,5=17,5 km
CHASE/COLLISION PROBLEMS
Solved example #3. Two objects with the same starting point moving with the same orientation, but starting at different times.
There has been a robbery and the thief escapes at a speed of 90 km/h. Six minutes later the police arrive at the bank and start chasing him at a speed of 120 km/h. Calculate when and where the police will catch the thief.
You set the stopwatch when the last one starts moving, in this case, the police. Therefore,
the thief has already covered certain distance that you have to calculate. Firstly, you have you change 6 min to hours:
In that time (0,1 h) the thief has moved forward e=v·t=90·0,1=9 km. When the police leave the bank, the initial position of the thief is 9 km.
1. We choose our reference frame and do a drawing.
2. We write the position-time equations for each moving object. Both of them have uniform motion in this case (xf = xi + v·t).
xThief = 9 + 90·t
xPolice = 0 + 120·t
3. When they find each other, xA = xB, so you can equal both equations to each other:
9 + 90·t= 0 + 120·t
You solve it: 9=30·t t=0,3 h
You can substitute this time in any of the two position-time equations (with one is enough):
xThief = 9 + 90·0,3=36 km xPolice = 0+120·0,3=36 km
The police catch the thief at a position of 36 km from where they started and 0,3 h later since they left (0,4 h after the thief escaped).
CHASE/COLLISION PROBLEMS
Solved example #4. Two objects with a different starting point moving with opposite orientation, but starting at different times. One of them has a uniform motion and the other one, a uniformly accelerated motion.
Two moving objects leave at the same time two different cities A and B separated by 5 km and they go to find each other. The object that leaves A has a constant speed of 20 m/s. The object that leaves B goes towards A with an initial speed of 30 m/s, but it brakes with an acceleration of 0,1 m/s2.
a) Calculate there and when they will find each other.
b) Draw an approximate position-time graph that represents the situation. 1. We choose our reference frame and do a drawing
2. We write the position-time equations for each moving object. Object A has a uniform motion (xf = xi + v·t) and B, a uniformly accelerated motion ( ).
xA= 0 + 20·t
xB =
3. When they find each other, xA = xB, so you can equal both equations to each other: 0 + 20·t= -50·t+5000=0
t1= 112,7 s; t2= 887,3 s
You get two positive times. This is because they are going to find each other twice: when B moves forwards and when B goes backwards because, as it is moving to the left and it brakes, there is going to be a moment when it stops and starts moving to the right, accelerating.
Now you can substitute this value of time in any of the position-time equations (with one is enough):
