#### Transcript Chapter 2

Chapter 2 Motion Along a Straight Line Motion Along a Straight Line 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 For constant acceleration we will develop the equations that give us the velocity and position at any time. we will study the motion under the influence of gravity close to the surface of the Earth. 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 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 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. x = xf-xi When an object changes its position, it has a velocity Check Point 2-1 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 Ans: b and c Ch 2-4 Average Velocity, Average Speed 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) t0 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 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 = 3t-2 2) x=-4t2-2 3) x= 2/t2 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=-8t m/s 3) v = -4/t3 4) v=0 a) 1 and 4 c) 2 and 3 Ch 2-6 Acceleration When an object changes its velocity, it undergoes an acceleration Average acceleration aavg aavg = v/ t = (vf-vi) /(tf-ti) Instantaneous acceleration a: a= lim (v/ t) t0 = dv/dt=d2x/dt2 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. If the sign are opposite, the speed decreases. Check Point 2-4 A wombat moves along an x axis. What is the sign of its acceleration if it is moving: a) in the positive direction with increasing speed, b) in the positive direction with decreasing speed c) in the negative direction with increasing speed, d) in the negative direction with decreasing speed? Ans: a) b) c) d) plus minus minus plus Ch 2-7 Constant Acceleration 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=v0+at x-x0=v0t+(at2)/2 v2=v02+2a(x-x0) x-x0=t(v+v0)/2 x-x0 =vt-(at2)/2 Check Point 2-5 The following equations give the position x(t) of a particle in four situations: 1) x=3t-4 2) x=-5t3+4t2+6 3) x=2/t2-4/t 4) x=5t2-3 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 = d2x/dt2=0 2) a = d2x/dt2=-30t+8 3) a = d2x/dt2 = 12/t4-8/t2 4) a = d2x/dt2 = 10 Ans: 1 and 4 ( constant acceleration case) Ch 2-9 Free Fall Acceleration Free fall acceleration ‘g’ dde to gravity g directed downward towards Earth’s center along negative y-axis with a = -g = -9.8 m/s2 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 = −9.8m/s2 Ch 2-10 Graphical Integration in Motion Analysis Integration of v-t graph to obtain displacement x x =x-x0=vt =v dt but v dt= area between velocity curve and time axis from t0 to t Integration of a-t graph to obtain velocity v Similarly v =v-v0= a dt a dt = area between acceleration curve and time axis from t0 to t