A range in a national park is driving at 47 km/h when a deer jumps onto the road 70m ahead of the vehicle . After a reaction of t seconds, the ranger applies the brakes to produce an acceleration of -2.00 m/s^2. What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?
To solve this problem, we can use the equations of motion.
Let's analyze the situation step by step:
1. First, let's find the time it takes for the ranger to reach the position of the deer if they were to continue moving with a constant speed of 47 km/h.
Distance = Speed * Time
We convert the speed from km/h to m/s:
47 km/h = (47 * 1000) m/3600 s = 13.06 m/s
Thus, the initial distance between the ranger and the deer is 70 m.
70 m = 13.06 m/s * Time
Now we can solve for Time:
Time = 70 m / 13.06 m/s
Time ≈ 5.36 s
2. Next, we need to calculate the distance the ranger will cover during the reaction time t.
Since the ranger is traveling at a constant speed, the distance covered during the reaction time is given by:
Reaction Distance = Speed * Reaction Time
Reaction Distance = 13.06 m/s * t
3. Finally, to avoid hitting the deer, the ranger needs to bring the vehicle to a stop before reaching the deer. To do this, the ranger applies the brakes to produce an acceleration of -2.00 m/s².
Using the equation of motion:
Final Velocity^2 = Initial Velocity^2 + 2 * Acceleration * Distance
Since the final velocity is 0 (as the ranger comes to a stop), the equation becomes:
0 = (13.06 m/s)^2 + 2 * (-2.00 m/s²) * Distance
Simplifying:
Distance = (13.06 m/s)^2 / (2 * 2.00 m/s²)
Distance ≈ 44.02 m
4. Now, to avoid hitting the deer, the ranger needs to stop the vehicle before reaching a distance of 70 m. Therefore, the reaction distance should be less than 70 m - 44.02 m.
Reaction Distance < 70 m - 44.02 m
13.06 m/s * t < 25.98 m
Dividing by 13.06 m/s on both sides:
t < 25.98 m / 13.06 m/s
t < 1.99 s
Therefore, the maximum reaction time allowed for the ranger to avoid hitting the deer is approximately 1.99 seconds.