{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

h1_s11

# h1_s11 - Phys 2101 Homework 1 Solution Spring `11 These...

This preview shows pages 1–3. Sign up to view the full content.

Phys 2101 Homework 1 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. (a) Denoting the travel time and distance from San Antonio to Houston as T and D , respectively, the average speed is avg1 (55 km/h)( /2) (90 km/h)( / 2) 72.5 km/h D T T s T T which should be rounded to 73 km/h. (b) Using the fact that time = distance/speed while the speed is constant, we find avg2 / 2 / 2 55 km/h 90 km/h 68.3 km/h D D D D s T which should be rounded to 68 km/h. (c) The total distance traveled (2 D ) must not be confused with the net displacement (zero). We obtain for the two-way trip avg 72.5 km/h 68.3 km/h 2 70 km/h. D D D s (d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero. (e) In asking for a sketch , the problem is allowing the student to arbitrarily set the distance D (the intent is not to make the student go to an Atlas to look it up); the student can just as easily arbitrarily set T instead of D , as will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to ( t 1 , x 1 ) = ( T /2, 55 T /2) and the second having a slope of 90 and connecting ( t 1 , x 1 ) to ( T, D ) where D = (55 + 90) T /2. The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to ( T, D ). The graph (not drawn to scale) is depicted below: 2.P.2.20. (a) Taking derivatives of x ( t ) = 12 t 2 2 t 3 we obtain the velocity and the acceleration functions:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
v ( t ) = 24 t 6 t 2 and a ( t ) = 24 12 t with length in meters and time in seconds. Plugging in the value t = 3 yields (3) 54 m x . (b) Similarly, plugging in the value t = 3 yields v (3) = 18 m/s. (c) For t = 3, a (3) = 12 m/s 2 . (d) At the maximum x , we must have v = 0; eliminating the t = 0 root, the velocity equation reveals t = 24/6 = 4 s for the time of maximum x . Plugging t = 4 into the equation for x leads to x = 64 m for the largest x value reached by the particle. (e) From (d), we see that the x reaches its maximum at t = 4.0 s. (f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted into the velocity equation, gives v max = 24 m/s. (g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s. (h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at that time is readily found to be 24 12(4) = 24 m/s 2 . (i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)). Thus, v avg = 54 0 3 0 = 18 m/s . 3. P2.22. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices, the initial position of the man is taken as the origin and his direction of motion during 5 min t 10 min is taken to be the positive x direction. We also use the fact that x v t ' when the velocity is constant during a time interval t ' .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern