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problem_set_4_due_Mar_19

# problem_set_4_due_Mar_19 - Physics 120 Spring 2008 Problem...

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Unformatted text preview: Physics 120 Spring 2008 Problem Set 4 due Wednesday, March 19 Note: It is not necessary to use the problem solving framework for these problems, but be sure to show all your work and evaluate your results. Fx (N) 6 1. A 2.0kg object initially at rest at the origin is subjected to the timevarying force shown in the figure at right. What is the object's xvelocity at t = 4 s? 3 0 1 2 3 4 t (s) 2. You are driving along at 20 m/s, trying to change a CD and not paying close attention to where you are going. When you look up, you find yourself 45 m from a railroad crossing. And, just your luck, a train moving at 30 m/s is only 60 m from the crossing. (See the figure below.) In a split second, you realize that the train is going to beat you to the crossing and you don't have time to stop. Your only hope is to accelerate enough to cross the tracks before the train arrives. If your reaction time is 0.50 s, what minimum acceleration does your car need for you to avoid a most unpleasant collision? 60 m TRAIN 45 m CAR 3. Careful measurements have been made of Olympic sprinters in the 100meter dash. A quite realistic model is that the sprinter's velocity is given by v x ( t ) = b ( - e - ct ) 1 where t is in s, vx is in m/s, and the constants b and c are characteristic of the sprinter. Sprinter Carl Lewis's run at the 1987 World Championships is modeled with b = 11.81 m/s and c = 0.6887/s. (A) According to this model, what was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s? (B) Find an expression for the distance traveled as a function of time t (i.e., find x(t)). Check that your expression is consistent with the initial condition x = 0 at t = 0. (C) Your expression from part b is a transcendental equation, meaning that you can't solve it for t. However, it's not hard to use trial and error to find the time needed to travel a specific distance. To the nearest 0.01 s, find the time Carl Lewis needed to sprint 100.0 m. If done correctly, it turns out that his official time was 0.01 s more than the time determined using this model, showing that this model is very good but not perfect. Note: For this problem, the following relations will prove useful: d a t a e = a e t where is a constant dt e a t 1 a dt = e t where is a constant a ...
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