Chapter 2 Motion Along a Straight Line Motion

Chapter 2 Motion Along a Straight Line

Motion Along a Straight Line n n In this chapter we will study kinematics, i. e. , how objects move along a straight line. The following parameters will be defined: Displacement Average velocity Average speed Instantaneous velocity Average and instantaneous acceleration n For constant acceleration we will develop the equations that give us the velocity and position at any time. n we will study the motion under the influence of gravity close to the surface of the Earth. n Finally, we will study a graphical integration method that can be used to analyze the motion when the acceleration is not constant.

Ch 2 -1 Motion Along a Straight Line n v v n n Motion of an object along a straight line Object is point mass Motion along a horizontal or vertical or inclined (line with finite slope) line Motion: Change in position No change in position, body at rest

Ch 2 -3 Position and Displacement v v v Axis are used to define position of an object Position of an object defined with respect to origin of an axis Position x of an object is its distance from the origin at any time t Displacement x, a vector, is change in position. v x = xf-xi When an object changes its position, it has a velocity

Check Point 2 -1 v v Here are three pairs of initial and final positions, respectively along x axis. Which pair give a negative displacement? a) -3 m, + 5 m b) -3 m, -7 m c) 7 m, -3 m § § Ans: x=xf-xi a) x=xf-xi=5 -(-3)=+8 b) x=xf-xi=-7 -(-3)=-4 c) x=xf-xi=-3 -(+7)=-10 ü Ans: b and c

Ch 2 -4 Average Velocity, Average Speed v v v v v Average Velocity vavg= x/ t vavg = (xf-xi) /(tf-ti)=Dispalcemnt/time Average speed Savg: a scalar Savg = total distance/ total time Instantaneous Velocity v: v= lim ( x/ t) t 0 Position-time graph used to define object position at any time t and to calculate its velocity v is slope of the line on positiontime graph

Check Point 2 -3 v v v v The following equations give the position x(t) of a particle in four situations ( in each equation , x is in meters, t is in seconds, and t>0): 1) x = 3 t-2 2) x=-4 t 2 -2 3) x= 2/t 2 4) x=-2 a) In which situation velocity v of the particle constant? b) In which v is in the negative direction? Ans: v=dx/dt 1) v=+3 m/s 2) v=-8 t m/s 3) v = -4/t 3 4) v=0 v v v a) 1 and 4 c) 2 and 3

Ch 2 -6 Acceleration v v v When an object changes its velocity, it undergoes an acceleration Average acceleration aavg = v/ t = (vf-vi) /(tf-ti) Instantaneous acceleration a: a= lim ( v/ t) t 0 = dv/dt=d 2 x/dt 2 Velocity-time graph used to define object velocity at any time t and calculate its acceleration a is slope of the line on velocity -time graph

Ch 2 -6 Acceleration If the sign of the velocity and acceleration of a particle are the same, the speed of particle increases. v If the sign are opposite, the speed decreases. v

Check Point 2 -4 v A wombat moves along an x axis. What is the sign of its acceleration if it is moving: v a) in the positive direction with increasing speed, v b) in the positive direction with decreasing speed v c) in the negative direction with increasing speed, v d) in the negative direction with decreasing speed? Ans: a) b) c) d) plus minus plus

Ch 2 -7 Constant Acceleration F F F Motion with constant acceleration has : Variable Slope of positiontime graph Constant Slope of velocity time graph Zero Slope of acceleration time graph For constant acceleration a =aavg= (vf-vi)/(tf-ti) vavg= (vf+vi)/2

Equations for Motion with Constant Acceleration v=v 0+at v x-x 0=v 0 t+(at 2)/2 v v 2=v 02+2 a(x-x 0) v x-x 0=t(v+v 0)/2 v x-x 0 =vt-(at 2)/2 v

Check Point 2 -5 q 1) 2) 3) 4) q The following equations give the position x(t) of a particle in four situations: x=3 t-4 x=-5 t 3+4 t 2+6 x=2/t 2 -4/t x=5 t 2 -3 To which of these situations do the equations of Table 2 -1 apply? Ans: Table 2 -1 deals with constant acceleration case hence calculate acceleration for each equation: 1) a = d 2 x/dt 2=0 2) a = d 2 x/dt 2=-30 t+8 3) a = d 2 x/dt 2 = 12/t 4 -8/t 2 4) a = d 2 x/dt 2 = 10 ü Ans: 1 and 4 ( constant acceleration case)

Ch 2 -9 Free Fall Acceleration v v Free fall acceleration ‘g’ dde to gravity g directed downward towards Earth’s center along negative y-axis with a = -g = -9. 8 m/s 2 equations of motion with constant acceleration are valid for free fall motion

Check Point 2 -6 a) What is the sign of the ball’s displacement for the ascent, from the release point to the highest point? B) What is it for the descent , from the highest point back to to the release point c) What is the ball’s acceleration at its highest point? a) (a) plus (upward displacement on y axis); (b) minus (downward displacement on y axis); (c) a = −g = − 9. 8 m/s 2

Ch 2 -10 Graphical Integration in Motion Analysis v v v Integration of v-t graph to obtain displacement x x =x-x 0=vt = v dt but v dt= area between velocity curve and time axis from t 0 to t Integration of a-t graph to obtain velocity v Similarly v =v-v 0= a dt = area between acceleration curve and time axis from t 0 to t

Old examples Which of the following statements is WRONG? A) A body can have constant velocity and still have a varying speed. Not Possible B) A body can have a constant speed and still have a varying velocity. uniform circular motion C) A body can have zero velocity and still be accelerating. free fall at max. height D) A body can have a constant acceleration and a variable velocity. by definition E) A body can have an upward velocity while experiencing a downward acceleration. free fall

Old examples n FIGURE 1 shows the motion of a particle moving along an x axis with constant acceleration. What is the magnitude of the acceleration of the particle? n n n n A) 4 m/s 2 B) 6 m/s 2 C) 3 m/s 2 D) 2 m/s 2 E) 5 m/s 2 Ans: x 0=x(0)= − 2 m x−x 0=v 0 t+ 12 at 2 x+2=v 0 t+ 12 at 2

Two cars are initially at rest. Car A is at x = 0, and car B is at x = + 600 m. They start to move, at the same time, along the same line in the positive x direction with constant accelerations: a. A = 4. 00 m/s 2 and a. B = 1. 00 m/s 2. How long does it take car A to overtake car B? A) 20. 0 s B) 30. 0 s C) 25. 0 s D) 34. 5 s E) 17. 5 s Ans: Let d 1=distance moved by A Let d 2=distance moved by B When they meet: d 1= d 2+ 600 �(1) d 2= v 0 t+ 12 a. Bt 2=12× 1. 00×t 2= 12 t 2 �(3) From (2)and (3), back to (1): 2 t 2=12 t 2+600 ⇒ 32 t 2=600 t 2= 23× 600=400 ⇒t=20. 0 s