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 ln 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 integration 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 particle 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 plane 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 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 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 position 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 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 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 vSWthe 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