Question

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 57 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.

A. How much distance is between you and the deer when you come to a stop?
B. What is the maximum speed you could have and still not hit the deer?

Answer 1

Step-by-step explanation:

Lets calculate the distance that the car has traveled first.

a)

as the car was traveling with the constant speed.

`v = (d)/(t)`

`d = v`×
`t`

d = (20)(0.5) = 10 m (1)

now we will calculate the distance that is required in order to stop the car.

`v^(2) _(f) = v^(2) _(i) + 2ad_(2)`

0 = (20)²+ 2(-10)
`d_(2)`

`d_(2)` = 20 m (2)

now by using (1) and (2) we can find out the total distance traveled y the car before it stops.

`d_(1) +d_(2) = 30 m`

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

`v_(f) = 0`

total distance traveled before stopping = 57 m

we must calculate the distance for reaction time with constant speed.

`d =v _(max) t = 0.5 v_(max)`

`d_(2) = 57 - d_(1)`

`d_(2) = 57 - 0.5v_(max)`

`v^(2) _(f) =v^(2) _(max) + 2ad_(2)`

0 =
`v^(2) _{max` + 2(-10)(57- 0.5
`v_(max)`)

`v^(2) _(max) = 1140 - 10v_(max)`

Apply the quadratic equation on it.

`v_(max)` = -b±
`\sqrt{b^(2)-4ac }/2a`

a = 1

b = 10

c = -1140

by putting these values we got

`v^{} _(max) = 29.13, - 39.13 m/s\\`

we neglect the negative value.

`v^{} _(max) = 29.13` m/s