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zubka84 [21]
3 years ago
6

You're driving down the highway late one night at 16.0 m/s when a deer steps onto the road 39.0 m in front of you. your reaction

time before stepping on the breaks is 0.50 s, and the maximum deceleration of your car is 12.0 m/s2 . how much distance is in between you and the deer when you come to a stop? what is the maximum speed you could have and still not hit the dear?
Physics
1 answer:
choli [55]3 years ago
5 0
The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a

x₂ = 0.5*a*t² = 0.5*v°²/a

The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7

You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀. 
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A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe
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Answer:

51.94 ft/s²

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Explanation:

t = Time taken = 4 s

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v = Final velocity

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34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s

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s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2

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