A driver in a car traveling at a speed of 70 mi/h sees a deer 90 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).
Vo = 70 mph = 102.7 ft/s = 31.3 m/s
The average speed while decelerating must be Vo/2 = 15.65 m/s
Deceleration time = 90 m/(avg. speed) = 5.75 s
Deceleration rate = Vo/(deceleration time) = 31.3/5.75 = 5.44 m/s^2
Check: distance travelled while decelerating =
Vo*t - (a/2)t^2 = 31.3*5.75 - (2.72)(5.75)^2 = 90 m
To calculate the minimum constant acceleration required for the car to stop without hitting the deer, we need to follow these steps:
Step 1: Convert the speed from miles per hour (mi/h) to meters per second (m/s). 1 mile is approximately equal to 1609.34 meters, and 1 hour is equal to 3600 seconds. So, the speed of the car in meters per second is:
70 mi/h * 1609.34 m/1 mi * 1 h/3600 s = 31.29 m/s (approximately)
Step 2: Use the following kinematic equation to find the minimum constant acceleration:
v^2 = u^2 + 2as
where:
v = final velocity = 0 (since the car needs to stop)
u = initial velocity = 31.29 m/s
a = acceleration
s = displacement = 90 m
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the values into the equation, we have:
a = (0 - (31.29 m/s)^2) / (2 * 90 m)
Step 3: Calculate the value of acceleration:
a = (-978.5041 m^2/s^2) / 180 m = -5.436 m/s^2 (to two decimal places)
The minimum constant acceleration required for the car to stop without hitting the deer is approximately -5.44 m/s^2. The negative sign indicates that the acceleration should be in the opposite direction of the car's initial motion.