a car is driving at 95 km/h down a highway when a deer runs infront of the car 80 m ahead. if the driver immediately applies the brakes and slows down at 4.2m/s^2, what happens?
95 km/h = 95,000m/3600s = 26.4 m/s.
V^2 = Vo^2 + 2a*d.
V^2 = 26.4^2 - 8.4*80 = 24.96,
V = 5 m/s. The car is still moving after traveling 80 m. Therefore, the car hits the deer!
95 km/h = 26.4 m/s
26.4 / 4.2 = stopping time
(26.4 + 0) / 2 = ave stopping speed
To analyze what happens when a car is driving at 95 km/h down a highway and a deer runs in front of it, we need to consider the car's initial velocity, the distance to the deer, and the car's deceleration due to braking.
1. Convert the car's initial velocity from km/h to m/s:
- 1 km/h = 1000 m/3600 s (conversion factor)
- 95 km/h * (1000 m/3600 s) = 26.39 m/s
2. Given that the deer is 80 meters ahead of the car, we can determine the time it takes for the car to reach the deer:
- Distance = Initial velocity * Time
- Rearranging the equation to solve for time: Time = Distance / Initial velocity
- Time = 80 m / 26.39 m/s ≈ 3.03 seconds
3. While the car is braking, it undergoes deceleration at a rate of 4.2 m/s². We can now determine how much the car will slow down during this time:
- Final velocity = Initial velocity + (Acceleration * Time)
- Final velocity = 26.39 m/s + (-4.2 m/s² * 3.03 s)
- Final velocity = 26.39 m/s - 12.726 m/s
- Final velocity ≈ 13.664 m/s
So, when the driver applies the brakes and the car slows down at a rate of 4.2 m/s², the car's final velocity will be approximately 13.664 m/s.