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Solution for (a)

1. Identify the knowns and what we want to solve for. We know that v 0 = 30 . 0 m/s size 12{v rSub { size 8{0} } ="30" "." "0 m/s"} {} ; v = 0 size 12{v="0 "} {} ; a = 7 . 00 m/s 2 ( a is negative because it is in a direction opposite to velocity). We take x 0 to be 0. We are looking for displacement Δ x , or x x 0 size 12{x - x rSub { size 8{0} } } {} .

2. Identify the equation that will help up solve the problem. The best equation to use is

v 2 = v 0 2 + 2 a x x 0 . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )} {}

This equation is best because it includes only one unknown, x size 12{x} {} . We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for x size 12{x} {} , but they require us to know the stopping time, t size 12{t} {} , which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for x size 12{x} {} .

x x 0 = v 2 v 0 2 2 a

4. Enter known values.

x 0 = 0 2 30 . 0 m/s 2 2 7 . 00 m/s 2 size 12{x - 0= { {0 rSup { size 8{2} } - left ("30" "." "0 m/s" right ) rSup { size 8{2} } } over {2 left ( - 7 "." "00 m/s" rSup { size 8{2} } right )} } } {}


x = 64 . 3 m on dry concrete . size 12{x="64" "." "3 m on dry concrete" "." } {}

Solution for (b)

This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is 5 . 00 m/s 2 size 12{ +- 5 "." "00 m/s" rSup { size 8{2} } } {} . The result is

x wet = 90 . 0 m on wet concrete . size 12{x rSub { size 8{"wet"} } ="90" "." "0 m on wet concrete" "." } {}

Solution for (c)

Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver’s reaction time.

1. Identify the knowns and what we want to solve for. We know that v - = 30.0 m/s ; t reaction = 0.500 s ; a reaction = 0 . We take x 0 reaction to be 0. We are looking for x reaction .

2. Identify the best equation to use.

x = x 0 + v - t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {} works well because the only unknown value is x size 12{x} {} , which is what we want to solve for.

3. Plug in the knowns to solve the equation.

x = 0 + 30 . 0 m/s 0 . 500 s = 15 . 0 m . size 12{x=0+ left ("30" "." "0 m/s" right ) left (0 "." "500 s" right )="15" "." "0 m"} {}

This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.

4. Add the displacement during the reaction time to the displacement when braking.

x braking + x reaction = x total size 12{x rSub { size 8{"braking"} } +x rSub { size 8{"reaction"} } =x rSub { size 8{"total"} } } {}
  1. 64.3 m + 15.0 m = 79.3 m when dry
  2. 90.0 m + 15.0 m = 105 m when wet
Diagram showing the various braking distances necessary for stopping a car. With no reaction time considered, braking distance is 64 point 3 meters on a dry surface and 90 meters on a wet surface. With reaction time of 0 point 500 seconds, braking distance is 79 point 3 meters on a dry surface and 105 meters on a wet surface.
The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time.


The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Source:  OpenStax, Kinematics. OpenStax CNX. Sep 11, 2015 Download for free at https://legacy.cnx.org/content/col11878/1.5
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