University Physics AI No. 1 Rectilinear Motion Class Number Name I． Choose the Correct Answer 1. An object is moving along the x axis with position as a function of time given by )(txx = . Point O is at 0=x . The object is definitely moving toward O when ( C ) (A) 0 d d < t x . (B) 0 d d > t x . (C) 0 d )d( 2 < t x . (C) 0 d )d( 2 > t x . Solution: If the object is moving toward O, the velocity and the position vector of the object have different direction. That is 0 d d <?= t x xxv , so the answer is C. 2． An object starts from rest at x=0 when t=0. The object moves in the x direction with positive velocity after t=0. The instantaneous velocity and average velocity are related by ( D ) (A) txtx /d/d < . (B) txtx /d/d = . (C) txtx /d/d > . (D) tx d/d can be larger than , smaller than, or equal to x/t. Solution: The instantaneous velocity and average velocity are both positive , the magnitude of them can not be compared. 3. An object is moving in the x direction with velocity )(tv x , and t v x d d is nonzero constant. With 0= x v when 0=t , then for 0>t the quantity tvv xx d/d is ( C ) (A) Negative. (B) Zero. (C) Positive. (D) Not determined from the information given. Solution: As 0= x v when 0=t , then for 0>t the velocity )(tv x has the same direction with t v x d d . So we have 0 d d > t v v x x , the answer is C. II. Filling the Blanks 1. The magnitude of the acceleration of a sports car that can drag race from rest to 100 km/h in 5.00s is 50/9m/s 2 or 5.56m/s 2 . Assume the acceleration is constant, although typically this is not a good assumption for automobiles. The ratio of the magnitude of this acceleration to the magnitude of the local acceleration due to gravity (g = 9.81 m/s 2 ) is 0.57 . Solution: According to the definition of the acceleration )m/s(56.5 9 50 5 3600/10100 2 3 == ? = ? ? = t v a The ratio is 57.0 81.9 9/50 = 2. The x-component of the position vector of a particle is shown in the graph in Figure 1 as a function of time. (a) The velocity component x v at the instant 3.0 s is -4m/s . (b) Is the velocity component zero at any time? yes If so, the time is 1.5s . If not, explain why not, . (c) Is the particle always moving in the same direction along the x-axis? No If so, explain what leads you to this conclusion. If not, the positions at which the particle changes direction is x=5m, t=1.5s . Solution: (a) According to the definition of the instantaneous velocity t tx v x d )(d = , tangent to graph at t=3s, so m/s4 5.1 6 ?=?= x v . (b) Because the tangent at t=1.5s is zero. (c) According to the tangent of the graph, the velocity is positive during t<1.5s and negative during t>1.5s 3. When a radio wave impinges on the antenna of your car, electrons in the antenna move back and forth along the antenna with a velocity component x v as shown schematically in Figure 2. Roughly sketch the same graph and indicate the time instants when (a) The velocity component is zero; a, c, e, g (b) The acceleration component is x a zero; b, d, f, h (c) The acceleration has its maximum magnitude. a, c, e, g Solution: (a) See the graph. (b)(c) According to the definition of the acceleration t tv a x x d )(d = , tangent to graph, we can drawing the conclusion. 4. The graphs in Figure 3 depict the velocity component x v of a rat in a one-dimensional maze as a function of time. Your task is to make graphs of the corresponding 2 1 3 0 2 4 t(s) x(m) Fig.1 v x t Fig.2 a b c d e f g h (i) Acceleration component x a versus time Solution: Using the definition of the acceleration t tv a x x d )(d = . (ii) The x-component of the position vector versus time. In all cases assume x=0m when t=0s. 1 2 3 4 0 2 -2 -1 1 a x (m/s 2 ) t(s) (a) 1 2 3 4 0 2 -2 -1 1 a x (m/s 2 ) t(s) (b) 1 2 3 4 0 2 -2 -1 1 a x (m/s 2 ) t(s) (c) 1 2 3 4 0 2 -2 -1 1 v x (m/s) t(s) (a) 1 2 3 4 0 2 -2 -1 1 v x (m/s) t(s) (b) 1 2 3 4 0 2 -2 -1 1 v x (m/s) t(s) (c) 1 2 3 4 0 2 -2 -1 1 v x (m/s) t(s) (d) 1 2 3 4 0 2 -2 -1 1 v x (m/s) t(s) (e) Fig.3 1 2 3 4 0 2 -2 -1 1 a x (m/s 2 ) t(s) (d) 1 2 3 4 0 2 -2 -1 1 a x (m/s) t(s) (e) Solution: Using 2 00 2 1 )( tatvxtx xx ++= and tavtv xxx += 0 )( , we have (a) ? ? ? ? ? <<? ≤≤ << = )s4s3(210 )s3s2(4 )s2s0(2 )( tt t tt tx (b) ? ? ? <≤+? <<? = )s4s2(42 )s2s0(2 )( 2 tt ttt tx (c) 2 2 1 2)( tttx +?= (d) ? ? ? ? ? <≤?? <<+? = )s4s2(6 2 1 4 )s2s0(2 )( 2 2 ttt ttt tx (e) ? ? ? ? ? <≤? << = )s4s2(4 )s2s0( 2 1 )( 2 tt tt tx III. Give the Solutions of the Following Problems 1 2 3 4 tHsL 1 2 3 4 xHmL 1 2 3 4 tHsL -4 -3 -2 -1 1 xHmL 1 2 3 4 tHsL -1 -0.5 0.5 1 1.5 2 mx 1 2 3 4 tHsL 0.5 1 1.5 2 xHmL 1 2 3 4 tHsL -2 -1.5 -1 -0.5 xHmL 1. Hang a mass on a vertical spring; imagine an origin at the place where the mass is at rest with i ? directed down as shown in Figure 4. Now set the mass into oscillation in the vertical direction by lowering the mass a distance A and letting it go. The subsequent motion is called simple harmonic oscillation and will be investigated in some detail in Chapter 7. We shall see that the position vector of a particle executing such one-dimensional oscillatory motion is given by the expression itAtr ? )]cos([)( ω= r , where A is expressed in meters, and ω is expressed in radians per second; both are constants. (a) Find the velocity vector and the acceleration vector as function of time. (b) What are the greatest magnitudes of the velocity and acceleration vectors? (c) What is the earliest (nonnegative) time that the position vector attains maximum magnitude? When the position vector has its greatest magnitude, what is the magnitude of the velocity vector? What is the magnitude of the acceleration vector at the same time? (d) At what time ( st 0≥ ) does the position vector first attain a magnitude of 0 m? At this time, what are the magnitudes of velocity and acceleration vectors? Solution: (a) The velocity vector is itAitA tt tr tv ? )]sin([ ? )]cos([ d d d )(d )( ωωω ?=== r v . The acceleration vector is itAitA tt tv ta ? )]cos([ ? )]sin([ d d d )(d )( 2 ωωωω ?=?== v v (b) AtAv ωωω =?= maxmax )]sin([ v AtAa 2 max 2 max )]cos([ ωωω =?= v (c) Using itAtr ? )]cos([)( ω= r , so when 1)cos( =tω , we have AtAr == maxmax )]cos([ ω . Thus ,...2,1,0=±= kkt πω and ,...2,1,0=±= k k t ω π Thinking the subsequent motion from beginning, so we neglect t=0s. The earliest time is )1( == kt ω π that the position vector attains maximum magnitude. At the same time, the magnitude of the velocity vector is m/s0)sin()sin()( =?=?= ω π ωωωω AtAtv v . The magnitude of the acceleration vector is Fig.4 x 0 0 A x i ? AAtAta 222 )cos()cos()( ω ω π ωωωω =?=?= v (d) Using itAtr ? )]cos([)( ω= r , so when 0)cos( =tω , we have 0)cos( == tAr ω . Thus ,...2,1,0 2 =±= kkt π π ω and ,...2,1,0 2 =±= k k t ω π ω π Then the time is )0( 2 min == kt ω π during s0≥t at which the position vector first attain a magnitude of 0 m. At this time, the magnitude of the velocity vector is m/s) 2 sin()sin()( AAtAtv ω ω π ωωωω =?=?= v . The magnitude of the acceleration vector is 222 sm/0) 2 cos()cos()( =?=?= ω π ωωωω AtAta v 2. One model for the motion of a particle moving in a resistive medium suggests that the speed decrease exponentially according to the expression t evtv β? = 0 )( , where 0 v is the speed of the particle when t = 0 s and β is a positive constant. (a) How long will it take the particle to reach half its initial speed? (b) What distance does the particle traverse during the time interval calculated in part (a)? (c) Through what distance does the particle move before it is brought to rest? Solution: (a) β β ββ 2ln 2ln 2 1 2 )( 0 0 =??=??=?== ?? tte v evtv tt (b) The distance that the particle traverse is βββ β β β β β 2 )1(dd)( 02ln0 ln2 0 0 ln2 0 0 ln2 0 v e v e v tevttvS tt =??=?=== ??? ∫∫ (c) According to the expression t evtv β? = 0 )( , When ∞→t , 0)( →tv . The distance the particle moves before it is brought to rest is βββ ββ 00 0 0 0 0 0 )10(dd)( vv e v tevttvS t-t- =??=?=== ∞ ∞∞ ∫∫