A deer runs into the road 1.24 km in front of a car. The driver, traveling at 70.5 km/hr, sees the deer and slams on the brakes which decelerates the car at a rate of 3.2 x 10-1 m/s2. Will the driver hit the deer?
Vo = 70.5km/h = 70,500m/3600s = 19.6 m/s.
V^2 = Vo^2 + 2a*d.
0 = 19.6^2 - 2*0.32*d, d = ?.
1.24 km = 1240 m.
The stopping distance(d) will be less than 1240 m and the driver stops before hitting the deer.
To determine whether the driver will hit the deer, we need to calculate the stopping distance of the car and compare it to the distance between the car and the deer.
First, let's convert the given units to match each other. We'll convert the car's speed from km/hr to m/s and the distance between the car and the deer from km to m.
Speed of the car = 70.5 km/hr = (70.5*1000 m) / 3600 s ≈ 19.583 m/s
Distance between the car and the deer = 1.24 km = 1.24 * 1000 m = 1240 m
Now, we can calculate the stopping distance of the car by using the formula:
Stopping distance = (initial velocity^2) / (2 * acceleration)
Given:
Initial velocity (vi) = 19.583 m/s
Acceleration (a) = -3.2 x 10^-1 m/s^2 (negative because the car is slowing down)
Stopping distance = (19.583^2) / (2 * -3.2 x 10^-1)
≈ 19.583^2 / -0.64
≈ 383.3349 / -0.64
≈ -598.961 m
The stopping distance is approximately -598.961 m.
Since the stopping distance is negative, it indicates that the car stopped before reaching the deer, which means the driver will not hit the deer.