Physics Help Forum Relay race

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 18th 2008, 11:06 PM #1 Junior Member   Join Date: Sep 2008 Posts: 11 Relay race in a relay race, runner A is carrying the baton and has a speed of 2.80 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.0800m/s^2. how long after wards will A catch up with B to pass the baton to B? What Im not sure is if this could be done as two problems. If so I believe I dont have enough information to do the first?
 Sep 18th 2008, 11:32 PM #2 Junior Member   Join Date: Apr 2008 Posts: 21 First, list what you do know: Runner A: $\displaystyle v_0 = 2.8 \frac{m}{s}$ $\displaystyle v_f = 2.8 \frac{m}{s}$ $\displaystyle x_0 = -25 m$ Because he is 25 m behind the start line. $\displaystyle x_f = ?$ $\displaystyle t = ?$ $\displaystyle a_A = 0 \frac{m}{s^2}$ Because the runner's velocity is supposed to be constant Runner B: $\displaystyle v_0 = 0 \frac{m}{s}$ $\displaystyle v_f = ?$ $\displaystyle x_0 = 0 m$ $\displaystyle x_f = ?$ $\displaystyle t = ?$ $\displaystyle a_B = 0.0800 \frac{m}{s^2}$ First, we use this basic kinematic equation: $\displaystyle x_f = x_0 + v_0t + \frac{1}{2}at^2$ For runner A: $\displaystyle x_f = 2.8t + \frac{1}{2}0t^2 - 25$ $\displaystyle x_f = 2.8t - 25$ For runner B: $\displaystyle x_f = 0 + 0t + \frac{1}{2}0.0800t^2$ If their final position is the same position, we can set their equations equal: $\displaystyle 2.8t - 25 = \frac{1}{2}0.0800t^2$ Now, we simplify: $\displaystyle 0.0400t^2 - 2.8t + 25 = 0$ Using the quadratic equation: $\displaystyle t = 59.5 s \ and \ 10.5 s$ I'm pretty sure that it's the 10.5... But if it is wrong, let me know and I will re-do it.
 Sep 18th 2008, 11:37 PM #3 Junior Member   Join Date: Sep 2008 Posts: 11 you are right. ok what I dont understand is why is Vf the same as Vo in runner A. Can you explain why?
 Sep 18th 2008, 11:40 PM #4 Junior Member   Join Date: Apr 2008 Posts: 21 If it doesn't specify a change in velocity or you can't find a reason for a change in velocity, you have to assume that it is constant. If the initial and final velocities were different and the change was not specified nor the final velocity given, then you would not be able to solve this problem at all. Last edited by Aryth; Sep 19th 2008 at 12:07 AM.
 Sep 18th 2008, 11:49 PM #5 Junior Member   Join Date: Sep 2008 Posts: 11 how long have you been doing physics problems cause you are really good. I have a hard time starting them once I get one well set up I could do it, but that is the hardest part. I always put out what I know but I end up having to many unknowns and then I don't know what to do?
 Sep 18th 2008, 11:59 PM #6 Junior Member   Join Date: Apr 2008 Posts: 21 You always list out the 6 basic kinematic quantities: $\displaystyle x_0 \ or \ y_0 = \text{Initial Position}$ $\displaystyle x_f \ or \ y_f = \text{Final Position}$ $\displaystyle v_{x0} \ or \ v_{y0} = \text{Initial Velocity}$ $\displaystyle v_{xf} \ or \ v_{yf} = \text{Final Velocity}$ $\displaystyle a_x \ or \ a_y = \text{Acceleration}$ $\displaystyle t = \text{Time}$ If you are missing more than 2 of these, your problem can't be solved using conventional kinematics equations. After you list those six in their appropriate directions, then you add any other variables, equations, or relations. Then you re-read the problem carefully, and see all the possible ways you could find the answer. Pick the one you think is the easiest and use it. If you are unsure about your answer, pick an alternative method and see if you get the same answer. If there are no alternative methods, just use your answer and try to generate something that is true that does not result from your answer. Last edited by Aryth; Sep 19th 2008 at 12:04 AM.

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### uniform acceleration problems on relay runner basis

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