Chapter-2 Motion Along a Straight Line

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

 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 = x f -x i  When an object changes its position, it has a velocity Ch 2-3 Position and Displacement

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=x f -x i a)  x=x f -x i =5-(-3)=+8 b)  x=x f -x i =-7-(-3)=-4 c)  x=x f -x i =-3-(+7)=-10  Ans: b and c

Ch 2-4 Average Velocity, Average Speed  Average Velocity v avg =  x/  t v avg = (x f -x i ) /(t f -t i )  Average speed S avg : a scalar S avg = 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 position-time graph

 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=-4t ) 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? Check Point 2-3  Ans: v=dx/dt 1) v=+3 m/s 2) v=-8t m/s 3) v = 2/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 a avg a avg =  v/  t = (v f -v i ) /(t f -t i )  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.  If the sign are opposite, the speed decreases.

 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? Check Point 2-4  Ans: a) plus b) minus c) minus d) plus

 Motion with constant acceleration has :  Variable Slope of position-time graph  Constant Slope of velocity - time graph  Zero Slope of acceleration - time graph  For constant acceleration a =a avg = (v f -v i )/(t f -t i ) v avg = (v f +v i )/2 Ch 2-7 Constant Acceleration

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

 The following equations give the position x(t) of a particle in four situations: 1) x=3t-4 2) x=-5t 3 +4t ) x=2/t 2 -4/t 4) x=5t 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 =-30t+8 3) a = d 2 x/dt 2 = 12/t 4 -8/t 2 4) a = d 2 x/dt 2 = 10  Ans: 1 and 4 ( constant acceleration case) Check Point 2-5

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/s 2  equations of motion with constant acceleration are valid for free fall motion

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 Check Point 2-6

 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  a dt = area between acceleration curve and time axis from t 0 to t Ch 2-10 Graphical Integration in Motion Analysis