Explanation:
Lets calculate the distance that the car has traveled first.
a)
as the car was traveling with the constant speed.
[tex]v = \frac{d}{t}[/tex]
[tex]d = v[/tex]×[tex]t[/tex]
d = (20)(0.5) = 10 m (1)
now we will calculate the distance that is required in order to stop the car.
[tex]v^{2} _{f} = v^{2} _{i} + 2ad_{2}[/tex]
0 = (20)²+ 2(-10)[tex]d_{2}[/tex]
[tex]d_{2}[/tex] = 20 m (2)
now by using (1) and (2) we can find out the total distance traveled y the car before it stops.
[tex]d_{1} +d_{2} = 30 m[/tex]
Distance between the car and the deer when it stops = 57 - 30 = 27 m
b)
lets calculate the maximum speed you could have and still not hit the deer.
we have
[tex]v_{f} = 0[/tex]
total distance traveled before stopping = 57 m
we must calculate the distance for reaction time with constant speed.
[tex]d =v _{max} t = 0.5 v_{max}[/tex]
[tex]d_{2} = 57 - d_{1}[/tex]
[tex]d_{2} = 57 - 0.5v_{max}[/tex]
[tex]v^{2} _{f} =v^{2} _{max} + 2ad_{2}[/tex]
0 = [tex]v^{2} _{max[/tex] + 2(-10)(57- 0.5[tex]v_{max}[/tex])
[tex]v^{2} _{max} = 1140 - 10v_{max}[/tex]
Apply the quadratic equation on it.
[tex]v_{max}[/tex] = -b±[tex]\sqrt{b^{2}-4ac }/2a[/tex]
a = 1
b = 10
c = -1140
by putting these values we got
[tex]v^{} _{max} = 29.13, - 39.13 m/s\\[/tex]
we neglect the negative value.
[tex]v^{} _{max} = 29.13[/tex] m/s