《西南交大 大学物理 英文 试题 答案no.a1-1.11348894》由会员分享，可在线阅读，更多相关《西南交大 大学物理 英文 试题 答案no.a1-1.11348894（6页珍藏版）》请在金锄头文库上搜索。
1、University Physics AI No. 1 Rectilinear Motion Class Number Name IChoose 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) 0ddtx. (C)0d)d(2tx. Solution: If the ob
2、ject is moving toward O, the velocity and the position vector of the object must be in different direction. That means 0dd. (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 e
3、qual. 3. An object is moving in the x direction with velocity )(tvx, and tvx ddis nonzero constant. With 0=xv when 0=t, then for 0t the quantity tvvxxd/d is ( C ) (A) Negative. (B) Zero. (C) Positive. (D) Not determined from the information given. Solution: If 0=xv when 0=t, then for 0t the velocity
4、 )(tvx has the same direction with tvx dd. We would have 0ddtvvx 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
5、 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. 5950 53600/1010023 =tva The ratio is 57. 081. 99/50= 2. Th
6、e 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 xv 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 alway
7、s 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 ttxvxd)(d=, it is the tangent of the graph at
8、t=3s. we can get m/s45 . 1 6=xveasily. (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 wit
9、h a velocity component xv 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 xazero; b, d, f, h (c) The acceleration has its maximum magnitude. a, c, e, g Solution: (a)
10、 See the graph. (b)(c) According to the definition of the acceleration ttvax xd)(d=, tangent to graph, we can draw the conclusion. 4. The graphs in Figure 3 depict the velocity component xv of a rat in a one-dimensional maze as a function of time. Your task is to make graphs of the corresponding 2 1
11、 3 0 2 4 t(s) x(m) Fig.1 vx t Fig.2 abc d e f gh(i) Acceleration component xa versus time Solution: Using the definition of the acceleration ttvax xd)(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
12、 -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 -
13、2 -1 1 ax(m/s) t(s) (e) (d)Solution: by using 2 0021)(tatvxtxxx+= and tavtvxxx+=0)(, we have (a) 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()(=AtAtvv. Fig.4 x 0 0 A x
14、 i The magnitude of the acceleration vector is AAtAta222)cos()cos()(=v(d) By using itAtr)cos()(=r, so when 0)cos(=t, we have 0)cos(=tAr. Thus ,.2 , 1 , 02=kkt and ,.2 , 1 , 02=kkt Then the time at which the position vector first attains the magnitude of 0 m is )0(2min=ktduring s0t. At this time, the
15、 magnitude of the velocity vector is m/s)2sin()sin()(AAtAtv=v. The magnitude of the acceleration vector is 222sm/0)2cos()cos()(=AtAtav2. One model for the motion of a particle moving in a resistive medium suggests that the speed decrease exponentially according to the expression tevtv=0)(, where 0v 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
