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1、Chapter 2 2 1 Air resistance acting on a falling body can be taken by the approximate relation for the acceleration agkv where k is a constant Derive a formula for the velocity of the body as a function of time assuming it stars from rest v 0 at t 0 Solution We find the function by integration 1 1 l
2、n 1 tv oo kt gkvdvdtdtdv gkv gkvg tve kkg 2 2 The acceleration of a particle is given by2at At t 0 v 10m s and x 0 a What is the speed as a function of time b What is the displacement as a function of time c What are the acceleration speed and displacement at t 5 0s Solution a We find the speed by i
3、ntegration 0 0 3 23 2 00 3 2 22 33 4 10 3 tv v dv aAt dt At dtdv vvA tvvA t tms b We find the displacement by integration 3 2 0 3 2 0 00 5 25 2 00 5 2 2 3 2 3 22 35 8 10 15 xt dx vvA t dt dxvA tdt xv tA tv tA t ttm c For the given time t 5 0s we have 2 4 5 25 80 ams vms xm 2 3 The position of a part
4、icle as a function of time is given by 2 7 608 85rtijt k m Determine the particle s velocity and acceleration as a function of time Solution We find the velocity and acceleration by differentiating 2 7 608 85 rtijt k 7 602vdrdtitk 2advdtk 2 4 At t 0 a particle stars from rest and moves in the xy pla
5、ne with an acceleration 2 4 03 0 aij m s Determine a the x and y components of velocity b The speed of the particle and c the position of the particle all as a function of time a We find the x and y components of velocity by integrating 00 4 0 4 0 4 x x x vt x x dv a dt dvdt vt 00 3 0 3 0 3 y y y vt
6、 y y dv a dt dvdt vt b The speed of the particle is 22 5 xy vvvt c We find the position by integrating 00 22 4 03 0 4 03 0 2 01 5 xy rt dr vv iv jtitj dt drtitj dt rt it j m 2 5 The position of a particle moving in the xy plane is given by 2cos32sin3 rt it j where r is in meter and t is in second a
7、Show that this represents circular motion of radius 2m centered at the origin b Determine the velocity and acceleration vectors as function of time c Determine the speed and magnitude of the acceleration d Show that the acceleration vector always points toward the center of the circle Solution The p
8、osition is 2 0sin3 02 0cos3 0rtitj a The magnitude of position is 22 2 0sin3 0 2 0cos3 0 2 0ttmr Thus the particle is always 2 0 m from the origin so it is traveling in a circle b We find the velocity and acceleration by differentiating 6 0sin3 06 0cos3 0 dr vtitj dt 18 0sin3 018 0cos3 0 dv atitj dt
9、 c The magnitude of velocity and acceleration are 22 6 0cos 3 0 6 0sin 3 0 6 0 vttm s 222 18 0cos 3 0 18 0sin 3 0 18 0 attm s d we see that 9 0ar so the acceleration vector is always opposite to the direction ofr and thus points toward the center of the circle 2 6 A swimmer is capable of swimming 1
10、00m s in still water a If she aims her body directly across a 75m wide river whose current is 0 80m s how far downstream from a point opposite her starting point will she land b How long will it take her to reach the other side Solution If vSBis the velocity of the swimmer with respect to the bank v
11、SWthe velocity of the swimmer with respect to the water and vWBthe velocity of the water with respect to the bank as shown in the diagram a We find the angle from SBSWWB vvv 0 0 80 tan0 80 1 00 38 7 WB SW v v Because the swimmer travels in a straight line we have tan75 0 8 60 shoreriver ddm b We can find how long it takes by using the components across the river 75 75 1 00 river SW d ts v v vWB v vSW v vSB
