Chapter 2 Describing Motion: Kinematics in One Dimension

Chapter 2 Describing Motion: Kinematics in One Dimension

Introduction (previous lecture) Reference Frames and Displacement (previous lecture) Average Velocity (previous lecture) Instantaneous Velocity (previous lecture) Acceleration (previous lecture) Motion at Constant Acceleration Falling Objects

Recalling Last Lecture You need to define a reference frame in order to fully characterize the motion of an object. In general, we use Earth as reference frame for our measurements. Displacement is how far an object is found from its initial position after an interval of time Δt: (2.1) Eq. 2.1 gives the magnitude of the vector x 1 x 2 Average velocity is given by: (2.4) And average acceleration is: (2.6) Both velocity and acceleration are vectors (give direction and magnitude) :

Acceleration, velocity, displacement, and time.

Deceleration

The car from the previous example, now moving to the left and decelerating.

Few Relevant Notes I. The magnitudes of the instantaneous velocity and instantaneous speed are the same, though this is not necessarily true for average velocity and average speed. II. If I give you an interval of time Δt = t 2 t 1 but do not specify the initial and final time t 1 and t 2, you can then assume t 1 = 0 such that Δt = t 2 t 1 = t 2 = t. III. Similarly, if I give you a displacement Δx = x 2 x 1 assume x 1 = 0 such that Δx = x 2 x 1 = x 2 = x. but x 1 is not specified, you can II and III are just a matter of redefining the origin of your coordinate system. For Example: 0 x 1 x 2 X 1 0 X 2 Translate the origin from 0 to x 1 by applying a transformation: X = x x 1 such that X 1 = 0 x 1 and X 2 = x 2 x 1 are the new set of coordinates.

Motion at Constant Acceleration It is not unusual to have situations where the acceleration is constant. In this lecture, we will see one of such situations experienced by you in your everyday life. But before, let s assume a motion in a straight line subject to a constant acceleration. In this case the following is valid: The instantaneous acceleration and the average acceleration have the same magnitude. We can then write: (2.8) We will use item II from the previous slide with following conventions: t 1 = 0, t 2 = t and also define the position and velocity at t 1 = t = 0 as x 1 = x 0, v 1 = v 0, x 2 = x v 2 = v for our future developments.

Motion at Constant Acceleration With this in mind, let s find some useful equations relating a, t, v, v 0, x and x 0 for cases with constant acceleration such that you can calculate any unknown variable if you know the others. Let s first obtain an equation to calculate the velocity v of an object subjected to a constant acceleration a: Using the definitions introduced on the previous slide, we can rewrite eq. 2.6 as: (2.9) Eq. 2.9 can be rearranged: (2.10) Eq. 2.10 allows you to calculate the velocity v of an object at a given time t knowing its acceleration a and initial velocity v 0.

Motion at Constant Acceleration Example: A car accelerates from rest at a constant rate of 5.0 m/s 2. What is its velocity after 5.0 s? Solution: Known variables: v 0 = 0.0 m/s a = 5.0 m/s 2 t = 5.0 s Using eq. 2.10:

Motion at Constant Acceleration We can also obtain an equation that gives the position x of an object knowing its constant acceleration a, initial position x 0 and the initial magnitude of its velocity v 0 : Using our definitions, eq. 2.4 can be written as: (2.11) On the other hand, since the magnitude of the velocity of the object changes from v 0 to v at a constant rate a, we have: (2.12)

But (Eq. 2.11) = (Eq. 2.12). Then: Motion at Constant Acceleration Using equation 2.10 in the above expression: or (2.13) Eq. 2.13 allows you to calculate the position of an object at a given time t knowing its acceleration a, the initial magnitude of its velocity v 0 and the initial position x 0.

Motion at Constant Acceleration There is another very useful equation that applies to motion at constant acceleration in situations where no time information has been provided. We will use equations 2.10 and 2.13 to eliminate the time variable: (2.10) (2.13) From eq. 2.10 Using this expression in 2.13: (2.14)

Motion at Constant Acceleration Example (Problem 28 from the text book): Determine the stopping distances for a car with an initial speed of 95 Km/h and human reaction time of 1.0 s, for an acceleration (a) a = -4.0 m/s 2 ; (b) a= -8.8 m/s 2? Solution: We want to convert from h to s. It is also wise to convert from Km to m: (remember to carry some extra figures through your calculations) Driver realizes he has to brake (v = v 0 ) Next, we have to define our reference frame (coordinate system) in order to calculate the position where the driver starts breaking. Driver starts braking (v = v 0 ) 0 x x 0 cars stops (v = 0)

Motion at Constant Acceleration Continuing Note that no information on the time needed to stop the car is provided. This problem is better approached with eq. 2.14. The position where the brakes are applied can be obtained if we observe that the car is moving at constant speed v 0 before the driver reacts. We can then use the equation of motion for constant speed to find x 0 : As we have already mentioned, the final speed v is zero in both (a) and (b) cases: v (final) = 0.0 m/s (continue on the next slide)

Motion at Constant Acceleration We have then the following initial conditions: (a) The stopping distance for a = -4.0 m/s 2 is: (b) Similarly, the stopping distance for a = -8.0 m/s 2 is:

Motion at Constant Acceleration Let s summarize the set equations we just found: (2.10) (2.12) (2.13) (2.14) Note that these equations are only valid when the acceleration is constant.

Motion at Constant Acceleration Example (Problem 21 from the text book): A car accelerates from 13 m/s to 25 m/s in 6.0 s. (a) What was its acceleration? (b) How far did it travel in this time? Assume constant acceleration. Solution: This problem is about choosing an appropriate equation from the set of equations we just found. For this purpose, first you have to write the parameters given in the problem: v 0 = 13 m/s v = 25 m/s t = 6.0 s (continue on the next slides)

Motion at Constant Acceleration (a) Here you want a knowing v 0, v and t. Which of the equations below you find more appropriate to solve this problem? (2.10) (2.12) (2.13) (2.14) It is clear that 2.10 is the better choice for this item. ;

Motion at Constant Acceleration (b) Here you want the distance travelled (x x 0 ) knowing v 0, v, t and a (calculated in item (a) ). Now, which of the equations below you find more appropriate to solve this problem? (2.10) (2.12) (2.13) (2.14) Either equation 2.13 or 2.14 is a good choice. Let s use 2.13:

Falling Objects We know from observation that an object will fall if we through it from a bridge. It will in fact gain speed as it falls. In other words, it is being accelerated. This acceleration is due to the action of the Earth gravitational force upon the object. We will come back to gravitation in few lectures from now. We use the letter g to denote the gravitational acceleration. g is always different from zero at any position but the very center of the Earth. We can say to very good approximation that g is constant: (2.15)

Falling Objects Some Notes: (a) It is also clear that certain objects fall faster than others even though we release them from rest (v=0) at the same time and from the same height. The reason for this is the air resistance. In fact, all objects experience the same gravitational acceleration g. (b) Note that we are dealing here with constant constant acceleration. Therefore all equations obtained on the previous slides apply to this case. (c) Since the motion will be vertical, we will use the y axis instead of the x axis.

Falling Objects Observing items (b) and (c) from the previous slide, we can rewrite the equations of motion as: (2.16) (2.17) (2.18) (2.19) Note: In most cases we will ignore the air resistance, unless otherwise stated.

Falling Objects Example (Problem 35 from the textbook): Estimate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before landing? Solution: Let s chose downwards as the positive y direction (again, this is just a convention). Again, we have to identify the initial conditions. 380 m g = 9.80 m/s 2 y - y 0 = 380 m +y v 0 = 0.0 m/s (continue on the next slides)

Falling Objects (a) Here you want t knowing v 0, g and (y y 0 ). Which of the equations below you find more appropriate to solve this problem? (2.16) (2.17) (2.18) (2.19) Equation 2.17 is more appropriated here: Here, the minus sign is not physical since it would imply King Kong traveling back time

Falling Objects (b) Here you want v knowing v 0, g, (y y 0 ) and t. Which of the equations below you find more appropriate to solve this problem? (2.16) (2.17) (2.18) (2.19) Equation 2.16 is more appropriated here: