Physics Help Forum hard kinematics question

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 10th 2009, 07:16 PM #1 Junior Member   Join Date: Sep 2009 Posts: 14 hard kinematics question You are arguing over a cell phone while trailing an unmarked police car by 25m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0s. At the beginning of that 2.0s, the police officer begins braking suddenly at 5.0 m/s^2. a) What is the separation between the two cars when you attention finally returns? Suppose that you take another 0.40s to realize your danger and begin braking. b) If you too brake at 5.0 m/s^2, what is your speed when you hit the police car? i converted the 110km/h to about 30.56m/s before i started the problem for part a), i just used the equation x = x0 + v0t + at^2 / 2. a = 5, t = 2, and v0 = 0 because the the 2 cars are going in the same direction so the relative velocity of the police car is just 0. plugging those values in, i got x - x0 = 10. since the displacement is 10, and the policeman was applying the brakes, he goes back 10m so the distance between the 2 cars is 25-10 = 15m now. this was my book's answer. for part b), i can't seem to get the right answer. i used the equations v = v0 + at and x-x0 = v0t + at^2 /2. i have x-x0 = 25, a = -5, and v0 = 0 since at time t=0, the two cars were traveling at the same speed in the same direction so i subtracted the two and got 0. when i solved, i got t = +/-3.1. so i plugged that into the first equation v = v0 + at and i got v = 30.56 + (-5)(3.1) and i got around 15 m/s. however the answer given by the book is 94 km/h. how did they get that answer?
 Sep 10th 2009, 07:43 PM #2 Physics Team     Join Date: Jul 2009 Posts: 310 Remember, the cop is still braking even after you start braking. __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC
Sep 10th 2009, 07:54 PM   #3
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 Originally Posted by Deco Remember, the cop is still braking even after you start braking.
how do i implement that information? do i say that they have the same acceleration in the same direction so the relative acceleration = 0?

 Sep 10th 2009, 08:05 PM #4 Physics Team     Join Date: Jul 2009 Posts: 310 You have to figure out the cops velocity after the intial 2.4 seconds, and therefore your relative velocity after the 2.4 seconds. $\displaystyle \frac {1}{2}a_ct^2 + 15 = \frac {1}{2}a_yt^2 + v_i t$ which simplifies to: $\displaystyle 15 = v_i t$ Solve for t and re-sub it into $\displaystyle v_f = v_i + at$ __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC
Sep 10th 2009, 09:02 PM   #5
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 Originally Posted by Deco You have to figure out the cops velocity after the intial 2.4 seconds, and therefore your relative velocity after the 2.4 seconds. $\displaystyle \frac {1}{2}a_ct^2 + 15 = \frac {1}{2}a_yt^2 + v_i t$ which simplifies to: $\displaystyle 15 = v_i t$ Solve for t and re-sub it into $\displaystyle v_f = v_i + at$
how did you get the first equation? and if you solve for t and substitute into $\displaystyle v_f = v_i + at$
won't you have 2 variables? final velocity and initial velocity?

 Sep 10th 2009, 09:07 PM #6 Physics Team     Join Date: Jul 2009 Posts: 310 The intial velocity is you car's velocity before you brake, 100 km/hr. As to how I got the first equation, the cop is 15 meters infront of you and is braking at the same rate you brake while you travel at a relative velocity to the cop. __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC
 Sep 10th 2009, 10:38 PM #7 Junior Member   Join Date: Sep 2009 Location: berkeley Posts: 19 I just had this same problem for homework last week. I'm no good at starting in the middle, so I'll just explain the logic from the start. a) the distance between the cars can be given as a function of time by $\displaystyle D(t)=x_{cop} - x_{me}$ $\displaystyle D(t)=(25 + 30.56t - 2.5t^2) - (30.56t)$, and for 2 seconds $\displaystyle D(2)=(25 + 30.56*2 - 2.5*2^2) - (30.56*2) = 15 (m)$ b) you need to get the velocity of the cop and distance D(t) at t=2.4 $\displaystyle v_{cop}(t)=30.56-5t$ $\displaystyle v_{cop}(2.4)=30.56-5*2.4=18.56$ $\displaystyle x_{cop}(t)=25 + 30.56t - 2.5t^2$ $\displaystyle D(2.4)=(25 + 30.56*2.4 - 2.5*2.4^2) - (30.56*2.4) = 10.6 (m)$ set both position functions equal to each other, subbing in the new numbers for the cop and your deceleration $\displaystyle x_{me} = x_{cop}$ $\displaystyle 30.56t - 2.5t^2 = 10.6 + 18.56t - 2.5t^2$ $\displaystyle 12t= 10.6$ $\displaystyle t=.8833$ plug that time into your velocity equation and you're good to go.

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