5.4 Breaking

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• Leerjaar 2

6.4 Braking and collisions

In traffic you always have to able to stop in time without having a collision.

Read paragraph 5.4 in complete silence

Write down the blue words, and write in your own words what they mean.

If you drive too fast, you might miss a curve.

If the initial speed is high, de braking distance is long. 

The greater the mass of the vehicle, the longer the breaking distance.

If the road is slippery you have to brake slowly.

If your braking force is to high the car skids.

And sometimes the brakes just fail....

initial speed and braking distance

If the speed is n times greater, 

the breaking distance becomes n² times longer.

v = 10 m/s

s = 5 m

v = 20 m/s

s = 20 m

v = 30 m/s

s = 45 m

n² = 2² = 4

n2 = 32 = 9

What's the STOPPING distance?

stopping distance = reaction distance + braking distance

Initial speed vs breaking distance

Speed is n times greater breaking distance is n^2 longer

example: v = 40 km/h  breaking distance 10 m 

40 x 2 = 80, so  breaking distance will be 2^2 = 4 times longer. 4 x 10 = 40 m 

A car moves at 40 km/h with breaking distance of 10 m. What is the breaking distance if it speeds up to 120 km/h.

1. 80
2. 90
3. 70
4. 40

40 x 3 =120 

3 ^3 = 9 so n=9 

10 x 9 = 90 m

An opossum is running at 2 m/s with a breaking distane of 1 m. What is the breaking distance if it accelerates to 4 m/s.

1. 12m
2. 8m
3. 2m
4. 4m

A teacher is riding his bicycle at 12 km/h with a breaking distance of 2m. Then he goes downhill at a speed of 60km/h. What is then the breaking distance?

1. 50
2. 75
3. 25
4. 15

A horse runs to the finish line at 39.6 km/h. The driver's reaction time is 0.2 s. The braking distance is 22.8 m. Determine the stopping distance.

A horse runs to the finishline at 39.6 km/h. The drivers reaction time is 0.2 s. The braking distance is 22.8 m. Determine the stopping distance.

given: v = 39.6 km/h = 11 m/s (39.6 / 3.6)

            t = 0.2 s

           s1 = 22.8m

asked: stopping distance

Solution : reaction distance : 

                   s2 = v x t = 11 x 0.2 = 2.2 m

           Stopping distance : s1 + s2 = 22.8 + 2.2 = 25m

A horse runs to the finishline at 39.6 km/h. The drivers reaction time is 0.2 s. The braking distance is 22.8 m. Determine the stopping distance.

given: v = 39.6 km/h = 11 m/s (39.6 / 3.6)
t = 0.2 s
s1 = 22.8m
asked: stopping distance
Solution : reaction distance :
s2 = v x t = 11 x 0.2 = 2.2 m
Stopping distance : s1 + s2 = 22.8 + 2.2 = 25m

Work

5.4: 1,2,3,4,5,6,7,8,9