First, here is a site talking a bit about a VERY exciting event happening Wednesday in Geneva, Switzerland. The largest particle super-collider (considered the most expensive science experiment ever built) will be turned on and tested tomorrow. This is a very big deal - read about it, and also check out this rap below that does a good job of showing what it will be able to do:
Solutions from the handout:
5.
a) 4 m/s2
b) Speeder displacement = 300 meters, police displacement = 200 meters
c) 100 meters
d) 20 seconds (They meet 10 seconds after the police officer stops accelerating, which is 10 seconds after the problem starts.)
6.
b) 4.5 seconds
c) The acceleration is the same at all three times, and has magnitude .267 m/s2
d) Area under the velocity vs. time graph during the acceleration = 1.2 meters
e) The total displacement during the trip is the total area under the curve. Since the area above the t-axis is the same as the area below, the net area is zero.
6(the second one)
a)5 m/s2
b) 1.25 m/s2
c) -2.5 m/s2
d) -5 m/s2
e) 0 m/s2
7.
a) Cart is at rest when velocity is zero, at t = 4 seconds, 18 seconds
b) The cart speeds up when the velocity and acceleration have the same sign. This occurs between t = 4 to 9 seconds, and t = 18 - 20 seconds
c) Displacement from t = 0 to t = 9 seconds is 1.6 meters - 2.5 meters = -0.9 meters, as found from the area above and below the t-axis. Since this is the displacement, and the displacement is x - x0, x = displacement + x0 = 2 meters + -0.9 meters = 1.1 meters.
8. velocity before entering the water is 8.85 m/s, which becomes v0 for the acceleration in the water. Applying v^2 = v0^2 + 2a*delta_x, a = 19.58 m/s2.
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