Lecture 2 Motion in one dimension Covers Sections 2 1 2 4 Mechanics is the study of motion kinematics and of the forces which produce motion dynamics We shall first study kinematics in one dimension To describe kinematics in one dimension we will introduce several different quantities We all have a colloquial understanding of terms like speed velocity acceleration distance The physics usage of these terms is similar to the colloquial usage but there are important differences These differences will become very pronounced when we study motion in two dimensions The quantities we need to understand and to calculate with are i Position x displacement xif xf xi distance d ii Average velocity v instantaneous velocity v average speed instantaneous speed iii Average acceleration a and instantaneous acceleration a A scalar quantity is just a number and has no direction A vector quantity has a magnitude and a direction Distance and speed are scalars while position velocity and acceleration are vectors Position velocity and acceleration To define a position we have to first decide on our reference point or origin In general we have to define a reference frame as we will see in the case of two dimensional motion The position of point P is an arrow which points from the origin to P This arrow has a magnitude and a direction which is either positive or negative in this one dimensional case The displacement between two positions P and Q is an arrow which points from P to Q and we define xP Q xQ xP For example if xP 10 and xQ 12 xP Q xQ xP 2 Notice that if xP 10 and xQ 8 xP Q 2 indicating that the arrow points in the negative direction In kinematics we are interested in motion and a useful way of looking at motion is through motion diagrams or motion equations which show position and or velocity and acceleration as a function of time For example consider motion described by x 3 4t t2 1 where x is in meters and t is in seconds Note that to ensure that this equation is dimensionally correct the units of the number 3 are meters the units of the number 4 are m s and the units of the 1 prefactor of the t2 term 1 are m s2 When motion equations are written down these units are usually omited but they are implied and sometimes you will be asked to figure out what the units must be to ensure that the equation is dimensionally correct We can draw many different graphs of x vs t Are all of them physically reasonable One restriction is that at any given time the position is single valued because we can t be in two places at the same time However we can return to the same position at a later time as occurs in the motion described by Eq 1 From motion diagrams or motion equations we can find the displacement between two times ti and tf so that xif x tf x ti Now we are ready to talk about velocity Velocity is a vector so it can also be thought of as an arrow which has a magnitude and a direction The average velocity is the rate of change of position with time In mathematical terms this is x tf x ti xif v 2 tf t i tif where we have defined tif tf ti The instantaneous velocity is the special case where tif becomes very small In fact in the limit where this quantity goes to zero is the limit in which the ratio xif tif becomes a derivative and we enter the world of calculus Though we shall not use calculus we are carrying out similar calculations when we calculate the instantaneous velocity Lets look at the motion diagram corresponding to Eq 1 Some points to note i The average velocity in going from B to D is zero ii The average velocity in going from A to B is 3m s iii The average velocity in going from C to D is 1m s iv The instantaneous velocity at any position is the slope of the graph at that position The acceleration is the rate of change of velocity and it is also a vector so it can be thought of as an arrow with a magnitude and a direction In mathematical terms the average acceleration is then given by a v tf v ti vif tf t i tif 3 where v t is the instantaneous velocity at time t and we use tif tf ti as in Eq 2 The instantaneous acceleration is the special case where tif becomes very small Looking at the motion curve corresponding to Eq 1 note the following 2 i The acceleration is the change in the slope of the motion curve ii The change in slope is negative for the whole motion curve which means that Eq 1 describes a system which is accelerating in the negative x direction iii Draw a graph and or write an equation for the case where the change in slope is positive Distance and speed As remarked above the distance and speed are scalars To be frank there is quite of bit of confusion about the definition of these quantities even in introductory textbooks The case where confusion arises is illustrated by the example Quiz 2 1 in the text page 27 The total displacement is zero however the total distance is 200 yards In our calculations we will take the distance to be the total distance travelled Note that the magnitude of the displacement is the net distance which in this example is zero In a similar way we shall define the average speed as follows average speed total distance total time 4 Note the following about distance and speed i The distance is taken to be the total distance travelled ii Distance and speed are always positive iii The instantaneous speed is the modulus or magnitude of the instantaneous velocity iv The average speed is always greater than or equal to the modulus of the average velocity over the same time interval Example Considering running from the origin ie xi 0 directly to a position x 25m and then reversing direction and running back to the final position xf 11 in a total time of tif 3s The displacment xif 11 so the average velocity v if 3 67m s but the total distance travelled is 39m so the average speed is 39m 3s 13m s 3
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