AP Physics Chapter 2 Review - cf.edliostatic.com

1 AP PhysicsChapter 2 Review2 v= x t=21+66 54() 21+22 6()[]m2sec= 2msec v=dvdt=d21+22t 6t2()dt=22 12tv=v1sec+v3sec2=10 14()msec2= position of a particle moving along the x axis is given by x = (21 + 22t - t2) m, where t is in seconds. What is the average velocity during the time interval t = sec to t = sec?3 vf2=vi2+2a xvf2 vi22 x=a220msec()2 450msec() ()=a 550357msec2= bullet is fired through a board, cm thick, with its line of motion perpendicular to the face of the board. If it enters with a speed of 450 msec and emerges with a speed of 220 msec, what is the bullet's acceleration as it passes through the board?4 v=dvdt=12t 3t2a=dvdt=12 6tAt max v, set a = 0.

2 T=2sec x=62()2 23x= position of a particle moving along the x axis is given by x = - , where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?5 v=dxdt=24 6t20=24 6t2t=2sec x=24t 2t3()mx=48m 16mx= particle moving along the x axis has a position given by x = (24t - ) m, where t is measured in seconds. How far is the particle from the origin (x = 0) when the particle is not moving?Find the area under the line from t = 1 sec to t = 6 sec and add the results to x = is the velocity of a particle moving along the x axis as shown below. If x = m at t = sec, what is the position of the particle at t = sec?

3 Tsec() vmsec()At t = 1 sec, x = two red areas cancel out and the green area = -3m. 2m 3m= 1m7 vmsec() tsec()Area is equal to the v. v=126msec2()5sec()=15msec v5sec=15msec+15msec=30msec8 amsec2() t = 0, a particle is located at x = 25m and has a velocity of 15 msec in the positive x direction. The acceleration of the particle varies with time as shown in the diagram above. What is the velocity of the particle at t = sec? v = d t= =vi+vf22v vf= () a=vf vit= particle confined to motion along the x axis moves with constant acceleration from x = m to x = meters during a second time interval. The velocity of the particle at x = m is msec.

4 What is the acceleration during this time interval?10 vf2=vi2+2a xa=vf2 vi22 xa=80msec()2 40msec()22200m()a= automobile moving along a straight track changes its velocity from 40 msec to 80 msec in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time? x=vft 12at22 x vft()t2=a240m 10msec()2sec()()2sec()2=aa= seconds, a particle moving with constant acceleration along the x axis goes from x = 10 m to x = 50 meters. The velocity at the end of this time interval is 10 msec. What is the acceleration of the particle?12 x=12at22 xt2= () ()2=aa= electron, starting from rest and moving with a constant acceleration, travels cm in msec.

5 What is the magnitude of this acceleration? rocket, initially at rest, is fired vertically with an upward acceleration of 10 msec2. At an altitude of km, the engine of the rocket cuts off. What is the maximum altitude it achieves? vo=0 vo=0 vHighest=0We re going to find the velocity of the rocket when the engine runs out of fuel (quits), first. Then we will find the height the rocket achieves when moving under only the influence of gravity. vf2=vi2+2a yvf2=2a yvf=2a yvf=210msec2()500m()vf=100msec14 vo=0 vHighest=0 v=100msec A BNow we ll find distance B and then add distance B to distance A. vf2=vi2+2g y0=vi2+2g y vi22g= yB 100msec()22 ()= yB yB=510m H= yA+ yBH=1010m1516 The ball, after two seconds: vf=vi+gt=20msec+ 10msec2()2sec()=0 Now, the ball is starting down with an initial velocity of zero while the stone is starting up.

6 Hball= ball is thrown vertically upward with an initial speed of 20 msec. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 msec. At what height above the release point will the ball and stone pass each other? 20 Hs=12gt2 Hs=24t 12gt2 20m 20 12gt2=24t 12gt2 20=24t t= Hs=24t 12gt2Hs= () 1210msec2() ()2Hs= vo v14=18msec vTop=0 34H vf2=v142+2g34H()0=v142+2g34H()0=v142+g32 H()2v142()3g=H= object is thrown vertically and has an upward velocity of 18 msec when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

7 Vf2=vo2+2gH()0=vo2+2gH()vo=2gHvo= ()22m()vo= vot y= +20t 60=0 t= 20+202 () 60() stone is thrown from the top of a building with an initial velocity of 20 msec downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground? Hs=40t 60m 60 10t Hs= () ()2H= object is thrown downward with an initial (t = 0) speed of 10 msec from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40 msec. At what height above the ground will the two objects pass each other?

8 60 Hs()= 10t y= 10t y=103sec()+ ()2 y= rock is thrown downward from an unknown height above the ground with an initial speed of 10 msec. It strikes the ground seconds later. Determine the initial height of the rock above the ground. y=vot y+ + ()23sec=vovo= ball thrown vertically from ground level is caught seconds later by a person on a balcony which is 14 meters above the ground. Determine the initial speed of the ball. vo v23=25msec vTop=0 13H vf2=v232+2g13H()0=v232+2g13H()3v232()2g= H= object is thrown vertically upward such that it has a speed of 25 msec when it reaches two thirds of its maximum height above the launch point.

9 Determine this maximum height.