Chapter 2. Motion Along a Straight Line. Goals for Chapter 2. To study motion along a straight line To define and differentiate average and instantaneous linear velocity To define and differentiate average and instantaneous linear acceleration
Chapter 2 Motion Along a Straight Line
Goals for Chapter 2 • To study motion along a straight line • To define and differentiate average and instantaneous linear velocity • To define and differentiate average and instantaneous linear acceleration • To explore applications of straight-line motion with constant acceleration • To examine freely falling bodies • To consider straight-line motion with varying acceleration
Displacement, time, and the average velocity—Figure 2.1 • Figure 2.1 allows us to illustrate each parameter.
Check your understanding 2.1 • Each of the following automobile trips takes one hour. The positive x-direction is to the east. • A travels 50 km due east. • B travels 50 km due west • C travels 60 km due east, then turns around and travels 10 km due west • D travels 70 km due east. • E travels 20 km due west, then turns around and travels 20 km due east. • Rank the five trips in order of average x-velocity from most positive to most negative. • Which trips, if any, have the same average x-velocity? • For which trip, if any, is the average x-velocity equal to zero? 4, 1, 3, 5, 2 1, 3 5
example • Two runners start simultaneously from the same point on a circular 200. m track and run in the same direction. One runs at a constant speed of 6.20 m/s, and the other runs at a constant speed of 5.50 m/s. • When will the fast one first “lap” the slower one and how far from the starting point will each have run? • When will the fast one overtake the slower one for the second time, and how far from the starting point will they be at that instant? 286 s, 1770 m, 1570 m 572 s, 3540 m, 3140 m
Example - Walking 1/2 the time vs. Walking 1/2 the distance • Tim and Rick both can run at speed vrand walk at speedvw, with vw < vr. They set off together on a journey of distance D. Rick walks half of the distance and runs the second half. Tim walks half of the time and runs the other half. a) Draw a graph showing the positions of both Tim and Rick versus time. b) Write two sentences explaining who wins and why. c) How long does it take Rick to cover the distance D? d) Find Rick's average speed for covering the distance D. e) How long does it take Tim to cover the distance?
solution x D D/2 t ½ tTim tTim tRick • Tim wins because he takes short time to cover the same distance as Rick. a. c. d. e.
Average and instantaneous velocity—Figure 2.6 • Refer to Example 2.1 and Figure 2.6 of a systematic solution of several bodies in motion.
Average and instantaneous velocities in x-t graph Secant line – average velocity tangent line – instantaneous velocity
example The automobiles make a 5 hour trip over a total distance of 200 km. • Which car starts later? • When does A & B pass each other? • Which car reaches 200 km first? • Calculate average speed of A and B.
B x(t +Dt) A x(t) t + Dt t The Derivative…aka….The SLOPE! • Suppose an eccentric pet ant is constrained to move in one dimension. The graph of his displacement as a function of time is shown below. At time t, the ant is located at Point A. While there, its position coordinate is x(t). At time (t+Dt), the ant is located at Point B. While there, its position coordinate is x(t + Dt)
B x(t +Dt) A x(t) t + Dt t The secant line and the slope Suppose a secant line is drawn between points A and B. Note: The slope of the secant line is equal to the rise over the run.
The “Tangent” line READ THIS CAREFULLY! • If we hold POINT A fixed while allowing Dt to become very small. Point B approaches Point A and the secant approaches the TANGENT to the curve at POINT A. B B x(t +Dt) x(t +Dt) A A x(t) x(t) t + Dt t + Dt t t We are basically ZOOMING in at point A where upon inspection the line “APPEARS” straight. Thus the secant line becomes a TANGENT LINE.
The derivative Mathematically, we just found the slope! Lim stand for "LIMIT" and it shows the ∆t approaches zero. As this happens the top numerator approaches a finite #. This is what a derivative is. A derivative yields a NEW function that defines the rate of change of the original function with respect to one of its variables. The above example shows the rate of change of "x" with respect to time.
In most Physics books, the derivative is written like this: Mathematicians treat as a SINGLE SYMBOL which means find the derivative. It is simply a mathematical operation. The bottom line: The derivative is the slope of the line tangent to a point on a curve.
example • Consider the function x(t) = 3t +2; What is the time rate of change of the function (velocity)? • This is actually very easy! The entire equation is linear and looks like y = mx + b . Thus we know from the beginning that the slope (the derivative) of this is equal to 3. We didn't even need to INVOKE the limit because the ∆t is cancel out. Regardless, we see that we get a constant.
Example • Consider the function x(t) = kt3, where k = proportionality constant. What happened to all the ∆t's ? They went to ZERO when we invoked the limit! What does this all mean?
The MEANING? • For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 meters. • The derivative, however, tell us how our DISPLACEMENT (x) changes as a function of TIME (t). The rate at which Displacement changes is also called VELOCITY. Thus if we use our derivative we can find out how fast the object is traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12 m/s
dx = ? dt Example x = 5
Example x = t5 x = t-5 x = t
Example x = 4t5
Example x =2t5 + 3t-1
Chain rule If x is a function of f, and f is a function of t, so indirectly, x is a function of t: x(f(t)) Example
Class work • Find the derivatives (dx/dt) of the following function • x = t3 • x = 1/t = t-1 • x = (6t3 + 2/t)-2 • x = 16t2 – 16t + 4
Average velocity vs. instantaneous velocity Example • A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t) = αt2 – βt3, whereα = 1.50 m/s2 and β = 0.0500 m/s3. • Calculate the average velocity of the car for the time interval: t = 0 to t = 4.00 s; • Determine the instantaneous velocity of the car at t = 2.00 s and t = 4.00 s.
Follow the motion of a particle—Figure 2.8 • The motion of the particle may be described from x-t graph.
Questions • The graph above shows velocity v versus time t for an object in linear motion. Which of the following is a possible graph of position x versus time t for this object?
Test your understanding 2.2 • According to the graph • Rank the values of the particle’s x-velocity vx at the points P, Q, R, and S from most positive to most negative. • At which points is vx positive? • At which points is vx negative? • At which points is vx zero? • Rank the values of the particle’s speed at the points P, Q, R, and S from fastest to slowest. P R Q, S R, P, Q = S
v2 – v1 ∆v aav-x = = t2 – t1 ∆t 2.3 average and instantaneous acceleration • The average acceleration of the particle as it moves from P1 to P2 is a vector quantity, whose magnitude equals to the change in velocity divided by the time interval. Velocity describes how fast a body’s position change with time. Acceleration describes how fast a body’s velocity change, it tells how speed and direction of motion are changing.
example • A racquetball strikes a wall with a speed of 30 m/s. the collision takes 0.14 s. If the average acceleration of the ball during collision is 2800 m/s/s. what is the rebound speed?
Average acceleration Instantaneous acceleration
Average and instantaneous acceleration Example 2.3 • Suppose the x-velocity vx of a car at any time t is given by the equation: vx = 60 m/s + (.50 m/s2)t2 • Find the change in x-velocity of the car in the time interval between t1 = 1.0 s and t2 = 3.0 s. • Find the average x-acceleration between t1 = 1.0 s and t2 = 3.0 s. • Derive an expression for the instantaneous x-acceleration at any time, and use it to find the x-acceleration at t= 1.0 s and t = 3.0 s. • 4.0 m/s • 2.0 m/s2 • a = (1.0 m/s3)t; 1.0 m/s2; 3.0 m/s2
example • The position of an object as a function of time is given by x(t) = at3 – bt2 + ct - d, where a = 3.6 m/s3, b = 5.0 m/s2; c = 6 m/s; and d = 7.0 m (a) Find the instantaneous acceleration at t = 2.4 s. (b) Find the average acceleration over the first 2.4 seconds. • 42 m/s2 • 16 m/s2
example • The position of a vehicle moving on a straight track along the x-axis is given by the equation x(t) = t2 + 3t + 5 where x is in meters and t is in seconds. What is its acceleration at time t = 5 s? (2 m/s2 )
Finding acceleration on a vx-t graph • Average acceleration can be determined by v-t graph
Finding the acceleration—Figure 2.12 • A graph of and t may be used to find the acceleration. • Average acceleration: the slope of secant line. • Instantaneous acceleration: the slope of a tangent line at point.
Caution: The sign of acceleration and velocity a is in the same direction as v a is in the opposite direction as v v: pos a: neg. v: pos a: pos. v: neg. a: pos. v: neg. a: neg.
We can obtain an object’s position, velocity and acceleration from it v-t graph Neg. 0 Pos. 0 0 Pos. Pos.
Finding acceleration on a x-t graph On a x-t graph, the acceleration is given by the curvature of the graph. Curves up from the point: acceleration is positive straight or not curves up or down: acceleration is zero Curves down: acceleration is negative
Finding x, v, a in x-t graph Neg. 0 0 pos. 0 pos. 0 pos.
V t a t Example • The figure is graph of the coordinate of a spider crawling along the x-axis. Graph its velocity and acceleration as function of time.
Check your understanding 2.3 • Refer to the graph, • At which of the points P, Q, R, and S is the x-acceleration ax positive? • At which points is the x-acceleration ax negative? • At which points does the x-acceleration appear to be zero? • At each point state whether the speed is increasing, decreasing, or not changing. S Q P, R P: v is not change; Q: v is zero, changing from pos. to neg., first decrease in pos. then increase in neg., R: v is neg., constant; S: v is zero, changing from neg. to pos., first decrease in neg. then increase in pos.,
Derive equations for motion with constant acceleration Given: (assume t0 = 0) derive: • vx = vx0 + axt • x = x0 + vx0 + ½ axt2 • vx2 – vx02 = 2ax(x – x0) Homework – extra credit
a-t graph A horizontal line indicate the slope = 0, a = 0 Since ax = ∆v / ∆t; ∆v = ax∙∆t which is represented by the area. The area indicate the change in velocity during ∆t
Use the equations to study motorcycle motion • Refer to Example 2.4 and use the equations in a practical example illustrating a motorcycle and rider.
Study two bodies with different accelerations • Refer to Example 2.5 and use the equations in a practical example illustrating a motorcycle and its rider chasing an SUV.