Answered

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WILL MARK BRAINLIEST (real answers only please)


A driver on a rural road slams on the brakes to avoid hitting a cow and comes to a stop. Skid marks made while the brakes were applied are 280 feet long. If the tread on the tires produced a coefficient of kinetic friction of 0.3 with the road, how fast was the car moving? 1 m/s=2.237 mph.


35 mph

25 mph

45 mph

50 mph

Sagot :

lublana

Answer:

The car moving with speed 50 mph

Explanation:

We are given that

Distance, s=280 feet=85.3 m

1 feet=0.3048 m

Coefficient of kinetic friction=[tex]\mu_s=0.3[/tex]

1 m/s=2.237 mph

We have to find the speed of car moving.

Speed, v=[tex]\sqrt{2\mu_s gs}[/tex]

Where [tex]g=9.8ms^{-2}[/tex]

Using the formula

Speed of the car,v=[tex]\sqrt{2\times 0.3\times 9.8\times 85.3}[/tex]

Speed of the car,v=22.4 m/s

Speed of the car, v=[tex]22.4\times 2.237\approx 50 mph[/tex]

Hence, the car moving with speed 50 mph

Option  d is true.

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