# ps2 - P207 Fall 2007 Solutions Assignment 2 1(a For a...

This preview shows pages 1–4. Sign up to view the full content.

P207 - Fall 2007 Solutions Assignment 2 1 (a) For a constant acceleration a , and initial position and velocity zero, v ( t ) = at (1) and x ( t ) = 1 2 at 2 (2) Solve Equation 1 for time to get t = v/a (3) Then substitution of Equation 3 into Equation 2 gives x = 1 2 a parenleftBig v a parenrightBig 2 = 1 2 v 2 a (4) Finally solve Equation 4 for a . a = 1 2 v 2 x (5) In order to reach a velocity of 360 km/h in a distance of 1.8km, we have, using Equation 5 that the acceleration will be a = 1 2 360 2 1 . 8 (km / hr) 2 km = 3 . 6 × 10 4 km / hr 2 = 3 . 6 × 10 4 km / hr 2 parenleftbigg 1000 m / km (3600 s / hr) 2 parenrightbigg = 2 . 8m / s 2 = 2 . 8 m / s 2 9 . 8 m / s 2 = 0 . 29 g (b) From Equation 3 we see that it will take a time t = 360 km / hr 3 . 6 × 10 4 km / hr 2 = 10 2 hr = 36s 1

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

View Full Document
(c) Acceleration, velocity and displacement vs time are plotted below. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 5 10 15 20 25 30 35 displacement [km] time [seconds] 0 50 100 150 200 250 300 350 400 0 5 10 15 20 25 30 35 velocity [km/hr] 0 0.5 1 1.5 2 2.5 3 3.5 4 0 5 10 15 20 25 30 35 acceleration [10 4 km/hr 2 ] (d) The average acceleration of a car is Δ v Δ t = 100 km / hr (9 s)(1 / 3600)hr / s = 4 × 10 4 km / hr 2 = 100 km / hr( 0 . 001 m / km 3600 s / hr ) 9 s = 3 . 1 m / s 2 (as compared to 3 . 6 × 10 4 km / hr 2 = 2 . 8 m / s 2 for the jet) (e) Runway 26 appears to be about half the length of Runway 22 or x = 1 2 (2 . 1)km = 1 . 05 km. In order to determine the velocity after a uniform acceleration a for a distance x we solve Equation 5 for v and find that v = 2 ax = ( 2(3 . 6 × 10 4 km / hr 2 )(1 . 05 km) ) 1 / 2 = 275 km / hr 2. Chapter 2, Problem 85. The distance that the car travels before coming to a stop is x = v 0 ( t react + t slow ) + 1 2 at 2 slow . (6) 2
v 0 is the initial velocity. t react is the time it takes to apply the brake. a is the acceleration after the brake is applied, and t slow is the time it takes to come to a stop after the brake is applied. Since for t > t react ,

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

View Full Document
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