Solutions Manual for University Physics with Modern Physics 2nd Edition by Wolfgang Bauer and Gary

Solutions Manual for University Physics with Modern Physics 2nd Edition by Wolfgang Bauer and Gary D.Westfall

Bauer/WestfallU: niversity Physics, 2E

stopped, the final velocities are zevrof =,

0 . vf = +v0 at= ⇒0 −= t a

v0 . Therefore, the ratio of time taken

to stop is Ratio =time of car 1=                 −v0/−a0 = 1. So the ratio is one half. time of car 2 −v0/          1 a0          2

2

Herea andv are instantaneous acceleration and velocitya. I=f 0 andv ≠ 0 at timet, then at that moment the object is moving at a constant velocity. In other words, the slope of a curve in a velocity versus time plot is zero at timet. See the plots below.

               The direction of motion is determined by the direction of velocity. Acceleration is defined as a change in velocity per change in time. The change in veloci∆tyv, , can be positive or negative depending on the values of initial

and final velocities∆,  =v vf −vi . If the acceleration is in the opposite direction to the motion, it means that

the magnitude of the objects velocity is decreasing. This occurs when an object is slowing down.

                If there is no air resistance, then the acceleration does neoptendd on the mass of an object. Therefore, both snowballs have the same acceleration. Since initial velocities are zero, and the snowballs will cover the same distance, both snowballs will hit the ground at the same time. They will both have the sam.e speed

Acceleration is independent of the mass of an object if there is no air resistance.

Snowball 1 will return to its original position af∆tetr , and then it falls in the same way as snowball 2. Therefore snowball 2 will hit the ground first since it has a shorter path. However, both snowballs have the same speed when they hit the ground.

Chapter 2: Motion in a Straight Line

                           Make sure the scale for the dpilsacements of the car is correct. The length of the car is 174.9 in = 4.442 m.

Measuring the length of the car in the figure above with a ruler, the car in this scale is 0.80 ± 0.05 cm. Draw vertical lines at the center of the car as shown in thuerfeigabove. Assume line 7 is the origixn=( 0).

Assume a constant acceleratioan= a0 . Use the equationsv = v0 at+ andx = x0 + v t0 + (1/2)at2 . When the car has completely stoppedv,= 0at t =t0 .

0 =+v0 at⇒0 −= v0      at0

Use the final stopping position as the origxin=, 0at t =t0 .

0 = x0+ v t0 0+ 1 at02

2

Substitutingv0 = −at0 and simplifying gives

x0 −at02+ 12at02= 0 ⇒ x0 − 12at02= 0 ⇒ a= 2tx020