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Difficult physics problems. need help!

  1. Sep 27, 2007 #1
    **I've done various steps to theses problems and can't seem to figure them out. If someone can either give me steps how to do them or just work them out for me I'd greatly appreciate it. Thanks!

    1) An electron with initial velocity v0 = 2.6x10^5m/s enters a region 1.0 cm long where it is electrically accelerated (Figure 2-26). It emerges with velocity v = 4.70x10^6 m/s. What was its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode-ray tube, used in television receivers and oscilloscopes.)

    I ended up getting 1.10 E 15 m/s^2 but I'm not sure if that's correct. I used the Vf^2-Vi^2=2ad equation.

    2) You are arguing over a cell phone while trailing an unmarked police car by 36 m. Both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.5 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 2.5 s, the police officer begins emergency braking at 5 m/s2.

    a) What is the separation between the two cars when your attention finally returns?

    b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car?

    I found a similar problem online to this and just redid the work and changed my numbers but I'm stuck on part b. I'm not even sure if what I have for a is correct. My work is shown below:
    distance between car and police= 36m speed of cars v=110km/hr =110(5/18) = 30.55m/s
    after police car breaks: initial v=20.55m/s, final v= 0m/s, a=-5m/s^2 from V^2-V^2-2as
    0^2-30.55^2=2(-5)s s=933.64/10= 93.364 m

    velocity of police car after 2.5 secs of breaking v=v+at 30.55+(-5)2.5=18.05 m/s
    distance traveled in this time by police is S: (18.05)^2-(30.55)^2-2(-5)s =60.78m
    distance traveled by car in 2.5s=velocity x time: 30.55 x 2.5= 76.375m
    separation between 2 cars: (36+60.78)-76.375= 20.405m

    3) A car is driven east for a distance of 45 km, then north for 30 km, and then in a direction 35° east of north for 22 km. Sketch the vector diagram and determine the (a) magnitude and (b) angle of the car's total displacement from its starting point.

    Here I tried to break the 3 distances into horizontal and vertical, which I thought was working but I ended up forgetting about the angle they give and now I'm totally confused. I'm not exactly sure how to find the magnitude nor the total displacement. I know you usually add up the distances to get the displacement but I have a feeling thats not what you do for this problem. Mind showing me the steps to these problems? Thanks
  2. jcsd
  3. Sep 27, 2007 #2


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    Homework Helper

    That number looks fine. Believe it or not, electrons really can pick up accelerations that huge from fairly modest electric fields. That's because their masses are so tiny. The force involved is about 1.0x10^-15 N, so the electric field only needs to be about 6300 N/C (or V/m), which is pretty mild.

    That's certainly one way to do it (there are a few similar approaches). I got essentially the same thing (allowing for round-off).

    Draw a sketch of what the car's path looks like. You'll want to deal with the components of displacement first. How far north of the starting point in total has the car traveled? How far east in total? The relatively hard part is getting the components of that 35º east of north leg (for which you'll need a little trig).
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