A motorist traveling at 15m/s encounters a deer in the road 38m ahead.
If the maximum acceleration the vehicle's brakes are capable of is -7m/s/s, what is the maximum reaction time of the motorist that will allow him or her to avoid hitting the deer?
Answer in units of s.
I have no clue how to solve this. There is also a part two to this question, which states:
If his or her reaction time is 1.44214s, how fast will he/she be traveling when he/she reaches the deer?
Answer in units of s.
Help would be greatly appreciated! <3
To solve this problem, we need to consider the motion of the vehicle and the reaction time of the motorist.
First, let's determine the minimum stopping distance required to avoid hitting the deer. We can use the following equation of motion:
\(d = v_0t + \frac{1}{2}at^2\)
Where:
- \(d\) is the distance traveled by the vehicle
- \(v_0\) is the initial velocity of the vehicle (15 m/s)
- \(a\) is the acceleration (in this case, the negative maximum braking acceleration of -7 m/s²)
- \(t\) is the reaction time of the motorist
We want to find the maximum reaction time (\(t\)) that will allow the motorist to avoid hitting the deer. Since the deer is 38m ahead, the minimum stopping distance should be equal to or greater than 38m. Therefore, we have:
\(38m = (15m/s)t + \frac{1}{2}(-7m/s²)t^2\)
Now, we need to solve the above equation to find the value of \(t\).
Step 1: Rearrange the equation to bring all terms to one side:
\(\frac{1}{2}(-7m/s²)t^2 + (15m/s)t - 38m = 0\)
Step 2: Solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. In this case, since the equation is not easily factorable, we will use the quadratic formula:
\(t = \frac{-b ± √(b^2 - 4ac)}{2a}\)
For our equation, \(a = \frac{1}{2}(-7m/s²)\), \(b = 15m/s\), and \(c = -38m\).
Step 3: Substitute these values into the quadratic formula and solve for \(t\):
\(t = \frac{-15m/s ± √((15m/s)^2 - 4 \cdot \frac{1}{2}(-7m/s²)(-38m))}{2 \cdot \frac{1}{2}(-7m/s²)}\)
Now we can simplify and calculate the possible values of \(t\):
\(t = \frac{-15m/s ± √(225m²/s² - 2 \cdot (-7m/s²) \cdot (-38m))}{-7m/s²}\)
\(t = \frac{-15m/s ± √(225m²/s² - 532m²/s²)}{-7m/s²}\)
\(t = \frac{-15m/s ± √(-307m²/s²)}{-7m/s²}\)
Since we cannot take the square root of a negative value in this context, it means that there is no real solution for \(t\) that allows the motorist to avoid hitting the deer. We can conclude that the motorist cannot avoid hitting the deer with the given conditions.
For the second part of the question, it states that the reaction time of the motorist is 1.44214s. Since it is not possible to avoid hitting the deer, the motorist will collide with it regardless of the reaction time. Therefore, the question does not provide enough information to determine the speed the motorist will be traveling at when reaching the deer.