Slide 1

Chapter 3

Uniformly Accelerated Motion

Topic OutlineDistance and displacementSpeed and velocityAverage and instantaneous velocity / accelarationOne dimensional motion (Kinematic equations)Freely falling bodiesProjectile motionHorizontal range and maximum height of projectileRange of projectile

Distance and DisplacementIf a particle is moving, we can easily determine its change in position. The changes is represented in term of space in any common measurement unit.

homeschool

DistanceDistance is how far something travels along its path. It is a scalar quantity.

homeschool3 kmDistanceFor example, you walk to school and back every schooling day. Your distance travelled is 6 km.

homeschool3 km3 km6 kmDisplacementThe displacement of a particle is defined as its change in position. It is a vector quantity.If we take the previous situation back, your displacement is actually zero.

Initial positionschoolhomeFinal positionDisplacementDisplacement is given by:We use the Greek letter to denote a large change in a quantity, that are significantly big enough to be measured.From this definition we see that x is positive if xf is greater than xi and negative if xf is less than xi .

Speed, Velocity and AccelerationSpeed = how fast youre going

Velocity = how fast youre going in a certain direction

Acceleration = how fast your velocity is changing (in a direction)SpeedThe speed of a particle, a scalar quantity, is defined as the total distance traveled divided by the total time it takes to travel that distance.SI unit for speed is meters per second (ms-1).

SpeedExample: A swimmer travels one complete lap (back and forth) in a pool that is 50 meters long. The first leg is covered in 20.0 seconds, the second leg is covered in 25.0 seconds. What was her average speed for the lap?

VelocityThe velocity of a particle, a vector quantity, is defined as the total displacement divided by the total time it takes to travel.SI unit for velocity is also meters per second (ms-1), but the direction of the movement is stated.

Velocity

Example: A ball is thrown to the air until it reaches the height of 5 meter before it falls back to the persons hand. Time taken for it to reach the throwers hand is 10 second. What is the velocity of the ball?Average Speed and VelocityThe average speed and velocity is calculated using the same formula as shown on few previous slides.Some books use the term average more to show the huge changes made by the moving particles.

Exercise #1A car travelled 100 km/hour East for 5.00 seconds, then reversed, and moved West for 3.00 seconds at speed of 50.0 km/hour. Assuming that East points towards the positive direction, find: i) Average velocity ii) Average speed.AnswerFirst, find the displacement and the distance travelled by the car.Displacement: Blue line red lineDistance: Blue line + red line

100 km/h for 5 s50 km/h for 3 sAnswerFor blue line: Change km/h to m/s

For red line: Change km/h to m/s

AnswerAverage AccelerationThe average acceleration of the particle is defined as the change in velocity vx divided by the time interval t during which that change occurred:

Acceleration has dimensions of length divided by time squared. The SI unit of acceleration is meters per second squared (m/s2).

Instantaneous Velocity and SpeedOften we need to know the velocity of a particle at a particular instant in time, rather than over a finite time interval.For example, even though you might want to calculate your average velocity during a long automobile trip, you would be especially interested in knowing your velocity at the instant you noticed the police parked in front of you.Instantaneous SpeedThe instantaneous speed of a particle is defined as the magnitude of its instantaneous velocity.For example, if one particle has a instantaneous velocity of +25 m/s along a given line and another particle at -25 m/s along the same line, both have instantaneous speed of 25 m/s.Instantaneous VelocityInstantaneous velocity vx equals the limiting value of the ratio x/t as t approaches zero:

This is also equal to the gradient / slope of tangent at that particular time.

Exercise #2 Find the runner's instantaneous velocity at t = 1.00 s. Position and time for a runner.t(s)x(m)1.001.001.011.021.101.211.201.441.502.252.004.003.009.00Instantaneous AccelerationIn some situations, the value of the average acceleration may be different over different time intervals. It is therefore useful to define the instantaneous acceleration as the limit of the average acceleration as t approaches zero.

Instantaneous AccelerationThe instantaneous acceleration equals the derivative of the velocity with respect to time.

Exercise #3 The velocity of a particle moving along the x axis varies in time according to the expression vx = (40 - 5t2) m/s, where t is in seconds. Find the average acceleration in the time interval t = 0 to t = 2.0 s.Determine the instantaneous acceleration at t = 2.0 s.Answera)

b)

One Dimensional MotionIf the acceleration of a particle varies in time, its motion can be complex and difficult to analyze. However, a very common and simple type of one-dimensional motion is that in which the acceleration is constant.One Dimensional MotionWhen this is the case, the average acceleration over any time interval equals the instantaneous acceleration at any instant within the interval, and the velocity changes at the same rate throughout the motion.The equation for ax in the previous slide can be modified to (assuming constant ax):

This powerful expression enables us to determine an objects velocity at any time t if we know the objects initial velocity (Vxi) and its (constant) acceleration (ax).One Dimensional MotionOne Dimensional MotionBecause velocity at constant acceleration varies linearly with time, the average velocity can be written as:

We can now formulate the displacement of any object as a function of time.

One Dimensional MotionUsing equation:We can substitute it into equation:

Forming the following equation

One Dimensional MotionFinally, using equations:

and

we can obtain an expression for the final velocity that does not contain a time interval:

One Dimensional MotionResulting in the following formula:

The equation for final velocity without time interval is:

To Summarize

The expressions above can be used to solve any problem involving one-dimensional motion at constant acceleration. The choice of which equation you use in a given situation depends on what you know beforehand.To SummarizeAnother way to write the equations (which also known as the 4 kinetics equations).

Exercise #4A train started from rest and accelerates at 2.0 m/s2 for 20 seconds. It then travels at a constant speed for 10 minutes. It then accelerate again for 5.0 seconds at the rate of 1.0 m/s2, before slowing down at a rate of -2.5 m/s2 until it stops. a) What is its speed at time t = 20 seconds?b) What is the total distance travelled by the train for the entire journey?c) What is the average speed for the train for the entire journey?AnswerFirst things first, draw the velocity-time graph.Vxt (s)20620625???Answera) Speed at 20 s is:t (s)20620625?40 m/s?Vx

AnswerSpeed at 625 s is:

t (s)20620625?40 m/sVx45 m/s

AnswerTime taken for train to stop is:

t (s)2062062564340 m/sVx45 m/s

Answerb) Distance travelled = area under the graph.

t (s)2062062564340 m/sVx45 m/sAnswerc) Trains average speed is:

Free Falling ObjectIn the absence of air resistance, all objects dropped near the Earths surface fall toward the Earth with the same constant acceleration g (also referred as gravity) measured to 9.8 m/s2You could try this: Simultaneously drop a coin and a cannon ball from the same height. If the effects of air resistance are negligible, both will have the same motion and will hit the floor at the same time.such motion is referred to as free fall.Free Falling ObjectA freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion.Objects thrown upward or downward and those released from rest are all falling freely once they are released. Any freely falling object experiences an acceleration directed downward, regardless of its initial motion.Exercise #5 A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Determine:the time at which the stone reaches its maximum height.the maximum heightthe time at which the stone returns to the height from which it was thrownthe velocity of the stone at this instant, and the velocity and position of the stone at t = 5s

Projectile MotionAnyone who has observed a baseball in motion (or, for that matter, any other object thrown into the air) has observed projectile motion.The ball moves in a curved path, and its motion is simple to analyze if we make two assumptions:(1) the free-fall acceleration g is constant over the range of motion and is directed downward(2) the effect of air resistance is negligibleExercise #6A canon fires a shell at an angle of 45o to the horizontal with a speed of 35 m/s. What is the time taken to reach the maximum height of the trajectory?What is the maximum height of the trajectory?What is the range of the shell?AnswerWhen the shell reaches the maximum height, it means that the vertical velocity component, Vy has reached 0.

45oxyV = 35 m/s

Answer (cont.)b) The maximum height of the trajectory is equivalent to asking the total displacement in the y direction.

Answer (cont.)(c) The range of the shell is the horizontal distance travelled by the shell. One major assumption is made here: The total time for the shell to move horizontally is 2t because it is the time required for the shell to reach maximum height (t) and for it to fall back to the its original level (t)= t + t = 2t. Also, there is no ax component.