Peter's Physics Pages

An Introductory Physics Course with Peter Eyland
Lecture 15 (Kinematics)

In this lecture the following are introduced:
• Zeno's paradox
• Average and instantaneous speed
• Position vs time graph to speed vs time graph
• Speed vs time graph to position vs time graph
• Average and instantaneous acceleration
• Speed vs time graph to acceleration vs time graph
• Acceleration vs time graph to speed vs time graph
• Constant acceleration graphs and equations

Zeno's paradox of Achilles and the tortoise

In an imaginary race between a fast runner (Achilles) and a slow runner (a tortoise) the tortoise is given a head start. After the start Achilles runs at constant pace, and while he does that the tortoise moves at a constant but slower speed. When Achilles gets to where the tortoise was initially T0, the tortoise has moved ahead to T1. Achilles doesn't pause where the tortoise was ( T0), and runs to where the tortoise now is ( T1), but during that time the tortoise moves ahead (to T2), not as far as Achilles did but has still moved ahead. Achilles continues to where the tortoise now is ( T2), but during that time the tortoise has moved on and remains ahead (at T3).

Achilles doesn't pause where the tortoise was, and runs to where the tortoise now is, but during that time the tortoise moves ahead, not as far as Achilles did but the tortoise is still ahead.

This argument can repeat indefinitely with an infinite number of time intervals, so reasoning this way, there is no time when Achilles will catch up to the Tortoise. This we know is wrong from experience. Clearly there is a fault in the reasoning. The fault lies in a natural belief that the sum of an infinite number of time intervals will be an infinite time. Although it may not be obvious, the sum of an infinite number of time intervals can be a finite time. The mathematics of this are given at the end of the lecture.

Kinematics

Kinematics (or "cinematics" with a soft "c") means "moving pictures", so kinematics is how the various aspects of motion are described in time. To do this, mathematical points have to be uniquely specified both in space and time Rene Descartes (1596 to 1650) argued that with a single reference point and a reference direction every point in a plane (or in space) can be uniquely identified. The Cartesian co-ordinate system has a reference point (or "origin"), a reference direction "x", and one (or two) perpendicular axes, "y" (for a plane) and "z" (for three dimensional space). A point is space is specified by co-ordinates along each axis, as shown on the left. He continued with this approach and created what is known as Coordinate Geometry.

 To introduce time, the horizontal axis is taken to give sequential moments in time and the vertical axis represents a direction in space. Two space directions ("x" and "y") can be reasonably represented in a "three dimensional" diagram, but three space axes takes a little imagination! In the diagram on the right, the co-ordinates (t,x) or (x,t) at a point on the line gives the straight-line distance of an object from its origin along the "x" axis, at the moment in time "t". Definition of average speed

An object's average speed between two times is defined by how quickly the position changes during that time interval, i.e.  This will have the S.I. unit of m.s-1, however in traffic regulations, km/h is used. (Note that non-S.I. units have a "forward slash" and not an index).

Example 1
A car is driven from Sydney to Byron Bay, a distance of 910 km in 13 hours. Find the average speed for the trip. On the trip, the car travelled at 70 km/h at least on two occasions, but it would have varied a lot during those 13 hours. To deal with this, another concept is needed, the speed at an instant of time i.e. the instantaneous speed.

Definition of instantaneous speed

The instantaneous speed is the slope of the tangent to a position vs time graph at a particular time. By using the slope of the tangent at a point, we avoid the difficulty of producing a time interval at a single point in time. Notice the symbolism, v is instantaneous speed and the slope of the tangent is represented by dx/dt. In reading such equations it never pays to focus on the symbols. That is, don't think "vee equals dee ex over dee tee" always read it in your mind as the instantaneous speed is given by the slope of the tangent on the position vs time graph.

For example, putting scales on the axes of the diagram above, the speed at a particular time, e.g. 2s can be calculated. Finding the speed vs time graph from the position vs time graph.

By working out the slopes at every point on a position vs time graph, the corresponding instantaneous speed vs time graph can be constructed. In interpreting these graphs, we can see that, on the position vs time graph, the steeper the slope the higher the speed.

Example 2
Comment on the instantaneous speeds at the four points labelled. At t1 the speed is slow. At t2 the speed is fast. At t3 the speed is zero. At t4 there is reverse speed.

Finding the position vs time graph from the speed vs time graph.

If you start from home and drive at a constant speed of 60 km/h for 2 hours, you will be 120 km from home. The speed vs time graph is shown on the left. The effect of speed through time is to change the position. By re-arranging the definition of average speed,distance travelled = (average speed) × (time involved). This product represents an area. A position vs time graph can be drawn from the areas under a speed vs time graph. Blue areas become red vertical lines. Each of the small blue areas is 60 × 0.5 which gives a position change of 30km. Note that this is a change in position and not an absolute position. The total blue area gives a position change of 60 × 2 which is 120km, the height of the line at 2h.

Position vs Speed graphs summary The position vs time graph shows the position at any time as the curvey red line. By measuring slopes at each point (e.g. the blue sloping line) the speed can be measured at each time, (shown as the vertical blue line in the lower graph). By measuring areas on the speed vs time graph (e.g. the red area) the change in position can be shown as the vertical red arrow in the upper graph at the end of the area.

Example 3
An object changes its position as shown. Find the speed between 0s and 1s, and also at 3s.   Example 4
Starting +2m from its origin, an object's speed is given by the following graph. Find the position of the object at 5s. Between 0s and 1s: Change in position = area of triangle between 0 and 1 = ½ × 1 × 1 = 0.5m Between 1s and 3s: Change in position = area of rectangle between 1 and 3 = 2 × 1 = 2m Between 3s and 5s: Change in position = area of rectangle between 3 and 5 = 2 × 2 = 4m> Total distance travelled is 6.5 m. However the object starts at +2m so its position is +8.5 m from its origin.

The techniques of finding slopes (how quickly something changes in time or space) and finding areas (the effect of something through time or space) will now be extended to speed and acceleration.

Definition of average acceleration This will have the S.I. unit of m.s-2. Definition of instantaneous acceleration

 The instantaneous acceleration is the slope of the tangent to a speed vs time graph at a particular time. By working out the slopes at every point on a speed vs time graph, we can construct the corresponding instantaneous acceleration vs time graph. Instantaneous acceleration and the speed vs time graph At t1 the acceleration is low. At t2 the acceleration is large. At t3 the acceleration is zero and the speed is constant (but not zero). At t4 there is a reduction in speed.

Finding the speed vs time graph from the acceleration vs time graph.

The area under the acceleration vs time graph gives the effect of acceleration through time, i.e. it give the spped change.

Example 5
An object accelerates from zero speed at its origin and at a constant rate of 3 m.s-2. Find the speed and position as functions of time while it continues at this rate. The area under the acceleration time graph gives the effect of acceleration through time, i.e. the speed change Δv. Δv = final speed - initial speed = v - v0 = 3t This is the length of the vertical arrow at time t on the speed vs time graph. The speed at any time is given by v = v0 + 3t but the intial speed is zero, so v = 3t The change in position will be the effect of speed through time, i.e. the triangular area on the speed vs time graph. Δx = final position - initial position = ½×t×3t = 1.5·t2. but the initial position is zero, so x = 1.5·t2.

Position-Speed-Acceleration graphs summary. The blue slope line on the position vs time graph gives the instantaneous speed which is the blue vertical arrow in the speed vs time graph below it. The curvey blue line of the speed vs time graph comes from all the slopes in the graph above it. The black slope line on the speed vs time graph gives the instantaneous acceleration which is the vertical black arrow in the acceleration vs time graph below it. The curvey blue line of the acceleration vs time graph comes from all the slopes in the graph above it. The green area under the acceleration vs time graph gives the change in speed, which is the green vertical arrow in the speed vs time graph above it. The red area under the speed vs time graph gives the change in position, which is the red vertical arrow in the position vs time graph above it.

Example 6
A hot air balloon rises vertically from rest on the ground with a constant acceleration of 3 m.s-2. Find
(a) the speed reached after 7 s.
(b) the height reached after 12 s. (a) The area under the acceleration time graph gives the the speed change Δv. Δv = final speed - 0 = 3t The speed at 7s is given by v = 3×7 = 21 m.s-1. (b) The change in height will be the triangular area on the speed vs time graph. Δx = final position - 0 = ½×t×3t = 1.5·t2. The height at 12s will be x = 1.5×122 = 216 m.

Example 7
A car is at its origin at time t=0s. It then has a speed given by v = 5 + 10t m.s-1. Find the acceleration and position at 3s.

 The speed vs time graph is:  The rectangle represents the change in position that would have occured if the car continued at its initial speed.
The triangle represents the additional distance that was added by the acceleration.
Since the car started from the origin, the final position is simply the change in position.

Constant acceleration graphs and equations

When a body has constant acceleration "a", the following applies. Constant acceleration under gravity

The Earth's force of gravity pulls mass towards it centre. Near the earth's surface, gravity causes masses to accelerate downwards with a constant value of 9.8 m.s-2. Using "g" as the symbol for this uniform acceleration, the following applies. Summarising:  Instantaneous positions, speeds and accelerations. When a body has constant acceleration "a", the following applies. Further practice questions

1. A submarine cruising at 4 m.s-1 accelerates at 3 m.s-2 for 2 second. Find (a) the speed reached after the 2 s. (ans 10ms-1), and (b) the distance travelled while accelerating. (ans 14m)

2. A bicycle rider uniformly accelerates to 30 km.hr-1 from 6 km.hr-1 in one-half minute. Find (a) the average acceleration during this time. (ans 2/9m.s-2), and (b) the distance travelled during this time. (ans 150m)

3. The gravitational acceleration on Jupiter is estimated at 26 m.s-2. A lead ball falls from rest and acquires a speed of 36.7 km.hr-1 just before it hits the surface of Jupiter. Find (a) the time taken in the fall. (ans 0.39s), and (b) the height from which the lead ball fell to give it this speed. (ans 2m)

4. When an object is released on the moon it will accelerate towards the surface at 1.6 m.s-2. A hammer is released from rest at 5 m above the surface when the time is zero second. Find (a) the time it takes to hit the surface. (ans 2.5s), and (b) the speed just before it hits the surface. (ans 4m.s-1)

5. In an 80 km.hr-1 speed restriction zone, a motorist makes an emergency stop with the brakes immediately locking and the wheels sliding. The length of the skid marks on the road were 19.6 m. A police officer assumes that the maximum deceleration of the car would not exceed the acceleration of a freely falling body (i.e. 9.8 m.s-2) and "books" the motorist for speeding.
Using the police officer's assumption, find (a) how long it took for the car to come to rest. (ans 2s), and (b) if the motorist was exceeding the speed limit. (ans no, 19.6m.s-1 < 22.2m.s-1)

6. A train moving with constant acceleration travels between two points 60 m apart in 6 s. Its speed as it passes the second point is 16 m.s-1. Find (a) the speed of the train at the first point. (ans 4m.s-1), and (b) the acceleration during this time. (ans 2m.s-1), and (c) the distance before the first point where it was stationary. (ans 4m)

7. A Jumbo jet is taken to have a constant acceleration from rest when it takes off. It has 1.8 km of runway in which to take off and needs to reach a speed of 360 km.hr-1. Find (a) the minimum acceleration needed to take off. (ans 2.8m.s-1), and (b) the time taken during this minimum acceleration. (ans 36s)

8. A car with good brakes moving at 64.8 km.hr-1 stops in 39.6 m. The corresponding distance when moving at 86.4 km.hr-1 is 64.8 m. Assume that the reaction time of the driver is the same for both situations. Assume that the deceleration of the car is the same for both situations. Find (a) the deceleration of the car. (ans 6m.s-2), and (b) the reaction time of the driver. (ans 0.7s)

E1. A springbok sees a cheetah and runs away at a constant speed of 13 m s-1. When the springbok is 19 m away the cheetah accelerates (constantly) from rest towards it and reaches a speed of 19.4 m s-1 in 2 s. The cheetah then travels at a constant speed towards the springbok for 4 s. Find
(a) the acceleration of the cheetah during the first 2 s.
(b) the distance travelled by the cheetah in the 6 s.
(c) the distance between the springbok and the cheetah at the end of the 6 s.

E2. By extending its legs a flea can accelerate upwards withan average acceleration of 1200 m s-2! Assume humans could have this constant acceleration over 0.54 m to create a vertical take off speed. Find
(a) the speed at which they would take off from the ground.
(b) the vertical height that they would reach.

E3. The gravitational acceleration on the surface of Mars is estimated at 3.8 m.s-2. A lead ball falls from rest and acquires a speed of 34.2 km.hr-1 just before it hits the surface of Mars. Find
(a) the time taken in the fall.
(b) the height from which the lead ball fell to give it this speed.

E4. A Cessna 150 aircraft needs to reach a speed of 126 km.hr-1 in order to lift off the ground. The effective runway length is 245m. Assume a constant acceleration. Find
(a) the time taken to reach the lift-off speed.
(b) the minimum (constant) acceleration required to achieve lift-off on the runway.

E5. A car accelerates uniformly from rest along a straight road. It reaches a maximum speed of 108 km hr-1 in one minute, after which it travels with constant speed for 3 minutes. Then it comes to rest (with uniform deceleration) in 15 seconds.
(a) Sketch a speed vs time graph for the motion, labelling axes with quantities, SI units and relevant values.
(b) Calculate the total distance travelled by the car.
(c) Calculate the average speed of the car for the whole motion.

E6. A Porsche racing car accelerates from rest, uniformly at 12 ms-2 for 2 s, then maintains constant speed for 1 s, and then decelerates uniformly, coming to rest in a further 3 s.
(a) Sketch an acceleration vs time graph for the motion, labelling axes with quantities, SI units and relevant values.
(b) Calculate the speed of the car at 4 seconds from the start.
(c) Calculate the total distance travelled by the car in the 6 seconds.

E7. A train driver in a passenger train moving at 32 m s-1 rounds a corner and sees a freight train 50m ahead on the same straight track and moving in the same direction with a speed of 20 m s-1. The brakes are applied on the passenger train giving the passenger train a constant deceleration of 1.46 m s-2. The freight train continues at constant speed.
(a) Sketch graphs showing for both trains:- speed vs time and position vs time, and
(b) determine if there will be a collision.

E8. At the instant a traffic light turns green a car accelerates from rest at a constant rate of 2.2 m.s-2. At the same instant, a truck overtakes the car, traveling with a constant speed of 16.5 m.s-1. Find
(a) how far from the traffic light the car overtakes the truck, and
(b) how fast the car is traveling when it overtakes the truck.

Achilles and the Tortoise

The resolution of the paradox shows that the sum of an infinite number of time intervals can be a finite time.
While Achilles covers a distance D, the Tortoise covers a fraction of this distance fD where f is less than 1. After 3 iterations: Achilles started from zero position and has moved through the first distance D to the tortoise and two more Tortoise distances. A3 = D + fD + f2D The Tortoise started from the postion D and has moved three Tortoise distances. T3 = D + fD + f2D + f3D It is seen that the Tortoise is ahead by f3D.

It will take an infinite number of iterations for Achilles to be at the same distance as the Tortoise. Call this distance S = A = T = D + fD + f2D + f3D + f4D + ...
Taking out the common factor D:
S = D( 1 + f + f2 + f3 + f4 + ... )
Multiplying by f:
fS = D( f + f2 + f3 + f4 + ... ) = S - D
Re-arranging the outside equalities:
S - fS = D
S = D/(1 - f)

This result indicates that where Achilles catches up to the Tortoise depends on the head-start distance, D, and the fraction of that distance that the Tortoise travels in the time that Achilles covers the head-start distance. For example, if the Tortoise covers ½ the distance that Achilles does each time then Achilles will catch up to the Tortoise in twice the head-start distance. email Write me a note if you found this useful

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