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katrin [286]
3 years ago
14

2. 3. 4. 5. 6. 7. A(n)results when a vehicle loses part or all of its grip on the road. is a technique that can be applied when

trying to move your vehicle out of deep snow. The action of a vehicle's rear end sliding out to a side is called a(n) is a technique of reducing your speed as quickly as possible while maintaining control of your vehicle. Driving at a speed where the stopping distance of your vehicle is longer than the distance you can see with your headlights is called occurs when a tire loses road surface contact and rises on top of water. It is a(n)when your front tires begin 1 18 1 1 to plow and your vehicle is not responding, or not responding as quickly as it should, to a steering input.
Physics
1 answer:
dexar [7]3 years ago
7 0

Answer:

1) Skid

2)<u> </u>Rocking

3) Oversteer situation

4) Controlled breaking

5) Overdriving headlights

6) Hydroplaning

7) Understeer situation

Explanation:

1) A <u>skid</u> results when a vehicle loses part or all of its grip on the road

2)<u> Rocking</u> is a technique that can be applied when trying to move your vehicle out of deep snow

3) The action of vehicle's rear end sliding out to a side is called an <u>oversteer situation</u>

4) <u>Controlled breaking</u> is a technique of reducing your speed as quick as possible while maintaining control of your vehicle

5) Driving at a speed where the stopping distance of your vehicle is longer than the distance you can see with your headlights is called <u>overdriving headlights</u>

6) <u>Hydroplaning</u> occurs when a tire loses road surface contact and rises on top of water

7) It is an <u>understeer situation</u> when your front tires begin to plow and your vehicle is not responding or not responding as quickly as it should, to a steering point

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(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

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So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

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