#### • Class 11 Physics Demo

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#### • formula for deceleration

Question:And how do you determine 1) what length of cord to use 2) the intended maximum weight of the user 3) the maximum resultant acceleration so as to not cause blackout(2-4 g force) Assuming that g=9.81ms^-2 and the height between the river and the bridge is 194.

Answers:Following are a few physics basics, but nobody in his/her right mind will design or use a cord without expert advice. See the ref. Assuming you want the jumper to just graze the water surface, you would design the cord with the jumper's weight in mind. Treating it as an ideal spring-mass system, you have: PE = mgh = kh^2/2 This means the energy of the fall (mgh) will equal the energy stored in the spring (kh^2/2) at maximum extension. From this you derive k. This is oversimplified because actually the jumper will fall freely for some distance dh before the spring starts to stretch. Thus PE in the spring actually = k(h-dh)^2/2. When you have solved for k you can solve for max. acceleration which = max. F/m. Max. F = k(h-dh). One would theoretically determine the length and type of bungee cord from k and h-dh. It would take deep research to find the stiffness/length, and limits such as max. load and max. elongation, of available cords. (Manufacturers don't seem to make all these data readily available.) The people who outfit jumpers for a living are really the ones you should consult.

Question:I know that to find out the distance travelled by, lets say a car, in a speed time graph, you have to measure the area beneath the line. However, if the acceleration is irregular before the car reaches constant speed and the deceleration is so irregular that it looks like a snake, then how do you actually calculate the distance travelled by the car? And also supposing that the area under the line of the speed-time graph isn't divisible by shapes...

Answers:Assuming that the graph is continuous, you integrate to find the area under the curve. Judging by the nature of your question, and your additional info, you haven't studied calculus. If the graph is a straight line, then you can easily form a rectangle, or a triangle. If the graph is a curve, then you can still form these rectangles, but they need to be much smaller. In fact, we make them infinitely small. Then we calculate the area of each rectangle and add them together. This is what the definite integral is made to do.

Question:These are my average velocities in order: 11.7, 10.6, 10.5, 6.5, 4.0 and another set: 26.1, 36.3, 38.7, 32.0, 27.8 do they have constant deceleration? how do you know?

Answers:Alright well I don't have time to sit here and calculate values (sorry) but I can explain how you would figure it out. Velocity is the change in position and Acceleration is the change in velocity. For there to be a constant change in acceleration (whether it's acceleration or deacceleration) each respective change in velocity will have to be equal to the next respective change in velocity, hence a constant acceleration. Basically find the difference between the values as you go along and if the difference between the velocities as you go through each of them is the same then it is constant. For example, above you have 11.7, 10.6, and 10.5. Well the first change in velocity, 11.7 to 10.6 has a difference of 1.1 in velocity. The next two numbers, 10.6 and 10.5 only have a difference of 0.1 in velocity so clearly the difference between the velocities as you go along is changing. Hope this was of some help. Let me know if you need clarification on anything.

Question:Hi full explanations please if possible. I'm trying to learn here. Thanks 1. A child leaves school and dashes across a crossing in from of a car travelling at 20ms-1. the driver sees him at a distance of 10 m and has an above average reaction time of 0.4s and breaks capable of delivering a deceleration of 20ms-2. Calculate a) the distance the car moves before the driver reacts. b) the distance the car moves while braking c) the total stopping distance 2. A car moves at 20ms-1experience a haed on collision and stops after the bonnet crumples a distance of 0.5m. Calculate a.)The time to stop b,.)The deceleration 3. Repeat if the car is travelling at 40ms-1

Answers:Calculations and explanations: The car is moving at 20 m/s. This is our initial velocity (u). At the end of our observation, the car has stopped, so it's final velocity (v) will be 0. a) The car moves 20 metres in one second, so in 0.4 of a second, it will move 20*0.4 = 8 metres. Speed = distance/time. So, if we rearrange that, we get distance = speed*time. b) In kinematic equations, distance is represented by s. Don't know why, it just seems to be. :S :| Anyway, the formula we will use is s=(v^2 - u^2)/2a. v=0 u=20, 20^2 = 400 a= -20, 2*-20 = -40 So, -400 / -40 = 10 metres. For the total stopping distance, we'll add the two answers together. 10+8 = 18 metres. For the second one, we're given distance (0.5), and initial velocity (20). Again, at the end of our observation, the car has stopped, so our final velocity is 0. We'll use the formula t=s/((u+v)/2). So this will be 0.5 / (20/2), which is 0.05 seconds. For the deceleration, we'll use the a=(v-u)/t. This will be -20/0.05, which is -400 m/s. There you go, mate!