Computer Science Canada car decelarates phyiscs |
| Author: | hamid14 [ Sat Jan 22, 2011 12:57 pm ] |
| Post subject: | car decelarates phyiscs |
This question is really bugging me. It says the car is decelerates at 5m/s squared, how far will the car have travelled before it stops, including the distance during the reaction time ? The answer is 106.4m, i dont understand how to get that. |
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| Author: | Tony [ Sat Jan 22, 2011 1:12 pm ] |
| Post subject: | RE:car decelarates phyiscs |
You're missing information (initial velocity of the car). Once you have that, you can figure out the time it takes to stop, average speed, and thus -- distance travelled. |
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| Author: | TerranceN [ Sat Jan 22, 2011 1:14 pm ] |
| Post subject: | RE:car decelarates phyiscs |
Well, what is the reaction time and how fast is the car going initially? Lets call this Tr and Vi. There are two sections to this problem, the first is before the brakes are pressed, where the car is going at a constant speed, the second is while the car is accelerating to rest. so Tr = reaction time Vi = initial speed a = acceleration d1 = distance traveled in first section d2 = distance traveled in second section dt = total distance lets set a direction as positive, how about the direction the car is initially moving. This will therefore make the acceleration -5m/s^2. The total distance is just a sum of the distances traveled in the two different sections, so dt = d1 + d2 Now to find d1 and d2, lets start with d1. Since this is constant velocity, d1 = Vi * Tr The second part uses a more complicated equation Since we want to find when the car stops, let Vf = 0 Lets use the equation which excludes the time it takes to stop (cause we don't have that, nor are we trying to find that), so Vf^2 = Vi^2 + 2*a*d2 solved for d2 is d2 = (Vf^2 - Vi^2) / (2*a) Then you can just find the two distances and add them together. Hope that helps. |
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| Author: | hamid14 [ Tue Jan 25, 2011 12:47 pm ] |
| Post subject: | Re: car decelarates phyiscs |
thanks, i didnt see your post, but i finished the exam review lol, thanks anyways. |
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