Here is some help for some Chapter 1 problems. Note solutions are not always exact but you should be able to get correct answer using the methods I present.
Problem 22 from Chap 1.
77 m + 813.8 m = 890 m or something between 889.5 and 890.5 m
Sum of two possible minimum values is 76.5 + 813.75 = 890.25
Sum of two maximum values is 77.5 + 813.85 = 891.35 so all possible values lie within 889.5 and 890.5. If we had said our answer was 77+ 813.8 = 890.8m that would have meant something between 890.75 and 890.85 which would not have included all possible values.
Problem 36 from Chapt 1 [coasts seems to mean maintains speed although most of us slow down when we coast]. In any case velocities for the zeroeth through 8th second are 0, 5,10,15,20, 25,25,25,25 m/s. Distances for the same times are: 0,2.5,10,22.5,40,62.5,87.5,112.5,137.5 m. Note you can find distance during acceleration simply by taking velocity at that time and dividing by 2 to get average during interval from start to that time then multiply by the time. This is the same as delta x = (a delta t) /2 x delta t or
delta x = a/2 delta t ^2. For last 3 seconds just add 25 m to position at previous second.
Problem 37 from Chapt 1. For easy arithmetic assume g = 10 m/s^2.
Can do this lots of ways easiest is use (vf^2-vi^2)/2a = delta x and rewrite for vf
vf = sqrt(vi^2+2a delta x)= sqrt ( 0+2x(-) 10x(-)30) = sqrt 600 = 24.4 m/s call it 24 [ note both a and delta x are negative]
for picture:
v is zero at start
at one second velocity is 10 m/s down after 1 second at 5 m down from start
at two seconds v is 20m/s down at 20 m down
watermelon reaches bottom in about 30m/((0+24.4)/2) =2.44 sec call it 2.4
40. delta t = (sqrt ( vi^2+2adelta x) –vi)/a = .37 s [note the sqrt term is just vf which comes to 6.3 m/s here, and that a is negative and delta x is positive]
just to check v avg =(10+6.3)/2 = 8.15 m/s vavg x delta t = 8.3 x .37 = 3.0 m
picture shows a =10 m/s^2 down and v = 10 at 0sec, 9 at .1 sec, 8 at .2 sec, 7 at .3 sec and 6.3 at 3.7 sec at impact.
Thursday, September 10, 2009
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment