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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s2. Part A How much distance is between you and the deer when you come to a stop? Express your answer to three significant figures and include the appropriate units. ss = nothing nothing Request Answer Part B What is the maximum speed you could have and still not hit the deer? Express your answer to three significant figures and include the appropriate units. vv = nothing nothing

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boffeemadrid

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

[tex]\text{Distance}=20\times 0.5=10\ m[/tex]

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m[/tex]

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

[tex]\text{Distance}=u\times 0.5=0.5u\ m[/tex]

[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0[/tex]

[tex]u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s[/tex]

Maximum speed of the car by which it will not hit the deer is 25.1 m/s

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