西南交大 大学物理 英文 试题 答案No.A1-1.11348894.pdf

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 C 0 d d 2 t x Solution If the object is moving toward O the velocity and the position vector of the object must be in different direction That means 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 D tx d d can be larger than smaller than or equal to x t Solution Though the instantaneous velocity and average velocity are both positive their magnitudes may not be equal 3 An object is moving in the x direction with velocity tvx and t vx 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 If 0 x v when 0 t then for 0 t the velocity tvx has the same direction with t vx d d We would have 0 d d t v v x x so 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 s2 or 5 56m s2 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 s2 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 vx d d it is the tangent of the graph at t 3s we can get m s4 5 1 6 x veasily b Because the tangent of the graph at t 1 5s is zero c According to the tangent of the graph the velocity is positive during t1 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 azero 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 draw 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 vx 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 ax m s2 t s a 1 2 3 4 0 2 2 1 1 ax m s2 t s b 1 2 3 4 0 2 2 1 1 ax m s2 t s c 1 2 3 4 0 2 2 1 1 vx m s t s a 1 2 3 4 0 2 2 1 1 vx m s t s b 1 2 3 4 0 2 2 1 1 vx m s t s c 1 2 3 4 0 2 2 1 1 vx m s t s d 1 2 3 4 0 2 2 1 1 vx m s t s e Fig 3 1 2 3 4 0 2 2 1 1 ax m s2 t s 1 2 3 4 0 2 2 1 1 ax m s t s e d Solution by 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 0 we neglect t 0s The earliest time when the position vector attains the maximum magnitude is 1 kt At the same time the magnitude of the velocity vector is m s0 sin sin AtAtv v Fig 4 x 0 0 A x i The magnitude of the acceleration vector is AAtAta 222 cos cos v d By using itAtr cos r so when 0 cos t we have 0 cos tAr Thus 2 1 0 2 kkt and 2 1 0 2 k k t Then the time at which the position vector first attains the magnitude of 0 m is 0 2 min kt during s0 t At this time the magnitude of the velocity vector is m s 2 sin sin AAtAtv v The magnitude of the acceleration vector is 222 sm 0 2 cos cos 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 0 where 0 v is the speed of the particle when t 0 s and 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 2ln 2ln 2 1 2 0 0 tte v evtv tt b The d