**Figure 4.42** represents a racetrack with semicircular sections connected by straight sections. Each section has length *d*, and markers along the track are spaced *d*/4 apart. Two people drive cars counterclockwise around the track, as shown. Car *X *goes around the curves at constant speed *v*c, increases speed at constant acceleration for half of each straight section to reach a maximum speed of 2*v*c, then brakes at constant acceleration for the other half of each straight section to return to speed *v*c. Car *Y *also goes around the curves at constant speed *v*c, increases its speed at constant acceleration for one-fourth of each straight section to reach the same maximum speed 2*v*c, stays at that speed for half of each straight section, then brakes at constant acceleration for the remaining fourth of each straight section to return to speed *v*c.

(a) On the figures below, draw an arrow showing the direction of the net force on each of the cars at the positions noted by the dots. If the net force is zero at any position, label the dot with 0. Description of **Figure 4.43**:

- The first dot on the left center of the track is at the same position as it is on the Car X track.
- The second dot is just slight to the right of the Car X dot (less than a dash) past three perpendicular hash marks moving to the right.
- The third dot is about one and two-thirds perpendicular hash marks to the right of the center top perpendicular has mark.
- The fourth dot is in the same position as the Car X figure (one perpendicular hash mark above the center right perpendicular hash mark).
- The fifth dot is about one and two-third perpendicular hash marks to the right of the center bottom perpendicular hash mark.
- The sixth dot is in the same position as the Car Y dot (one and two third perpendicular hash marks to the left of the center bottom hash mark).

(b) i. Indicate which car, if either, completes one trip around the track in less time, and justify your answer qualitatively without using equations.

ii. Justify your answer about which car, if either, completes one trip around the track in less time quantitatively with appropriate equations.

a) see video

b) car Y takes less time. See video for derivation and explanation.

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View sample solution*d*over 4 because our figure here says that

*d*is the distance of four sections and so when we consider one straight section that's one quarter

*d*. There are two time intervals to consider for car X, the time interval during which it is accelerating up to its top speed here and then the time interval when it is decelerating from the mid-point to the end of the section. So, this distance here is half of

*d*over 4 which is

*d*over 8 and that's going to be the average velocity multiplied by the time. The average velocity will be

*vc*at the start plus two

*vc*we're told is the final velocity and all that divided by two and then multiplied by

*delta t x one*. Then collect those two like terms, you get three

*vc*times

*delta t x one*over two equals

*d*over 8, switching sides around as well there. Then multiply both sides by two over three

*vc*to solve for which will be

*d*over

*12 vc*.

*Delta t x two*is going to be a similar story. It's also

*d*over 8 for this section here and then it starts at two

*vc*is its initial speed at this midpoint, and ends at

*vc*at the end of this section, all divided by two and it's all the same number, just a different order of the terms there. So the answer will be the same as for

*delta t x one*. So the total time to do one section between perpendicular hash marks, is going to be

*delta t x one*plus

*delta t x two*which is basically two times

*d*over

*12 vc*which is

*d*over

*six vc*. So the total time around the entire track, will be

*two d*over

*vc*because there is a -- this distance

*d*here and then this distance

*d*here on the curved portions, and the time to do each of them is

*d*over

*vc*because it goes at constant speed

*vc*over each of these curved portions. There is two of them so two times

*d*over

*vc*plus eight of these little sections, straight sections that we just calculated the time for. So there's one, two, three, four, five, six, seven, eight, and then so that's eight times

*delta t x*. Substituting for

*delta t x*is

*d*over

*6 vc*. We have

*two d*over

*vc*plus

*four d*over

*three vc*then making a common denominator by multiplying top and bottom of this fraction by three, we get

*6d*over

*three vc*and that gives a total of

*ten d*over

*three vc*is the time for car X to complete one lap. Then we've car Y to consider and there are four sections, sorry, there is three time intervals to consider for one straight section. There is the time interval during which it is accelerating,

*delta t y one*, there's the -- and that occurs over a distance of

*d*over 16 so that's the one quarter mark of a straight section and a straight section is of distance

*d*over 4, so a quarter of

*d*over 4 is

*d*over 16. Then we have half of this straight section which is

*d*over 8, during which the car is going at constant speed, a top speed of

*two vc*. Then again we have another section

*delta t y three*when the car is slowing down from

*two vc*back to its speed of

*vc*. Okay. So, the total time for this straight section will be

*delta t y one*plus

*delta t y two*plus

*delta t y three*and so let's figure out where each of them are. So this distance here

*d*over 16, is going to be the average velocity during that time interval, multiplied by the time interval. So it starts at

*vc*and it ends at

*two vc*here and I'l' divide that sum by two, and then multiply that by

*delta t y one*and collect the like terms at the top to get

*three vc delta t y one*over two equals

*d*over 16. Then multiply both sides by two over

*three vc*and this gives

*delta t y one*is

*d*over

*24 vc*. This middle section here which has a distance of

*d*over 8, is going at constant speed and the speed is

*two vc*and that's getting multiplied by

*delta t y two*which we then solve for by dividing both sides by

*two vc*

*. This gives us**d*over*16 vc*is*delta t y two*.*Delta t y three*is the same numbers as for finding*delta t y one*but just a different order of the terms in the numerator there because it starts at*two vc*and it ends at speed of*vc*. So the` answer will be the same for*delta t y three**as it was for*and then getting a common denominator of*delta t y one*,*d*over*24 vc*. So adding those together, we have two times*d*over*24 vc*plus*d*over*16 vc*. That means*d*over*12 vc*plus*d*over 16 vc*48 vc*, we end up with*four d*over*48 vc**plus**three d*over*48 vc*which is*seven d**over**48 vc*. So that's the time for one straight section for car Y. Then the total time to do the entire lap for car Y, is going to be*two d*over*vc*which accounts for the constant speed curved portions just as we had for car X, plus eight times these straight section times. So that's eight times*seven d*over*48 vc*. So that makes*two d*over*vc*plus*seven d*over*six vc*and 48 divided by eight being six, and a common denominator of*six vc**is called for, so we have**12 d*over*six vc*plus*seven d*over*six vc*plus*seven d*over*six vc*which is*nineteen d*over*six vc*. So we need to compare that to*t x*now. So*t x*we found before is*10 d*over*three vc*, found that here, and to make the comparison easier let's make a common denominator. So that's*twenty d*over*six vc*when you multiply top and bottom by two. So we can see that the total time for car Y being*nineteen d*over*six vc*is less than the total time for car X which is*twenty d*over*six vc*.