you are riding in a car traveling at 60mi/hr in a 50 mi/hr zone. you spot a policeman 10 feet ahead. you slow down with an acceleration of -2ft/sec2 for 3 seconds. Do you get a ticket
Vo = 60mi/h * 5280Ft/mi * 1h/3600s = 88 Ft/s.
V = 50/60 * 88Ft/s = 73.3 Ft/s.
V = Vo-a*t = 88-2*3 = 82 Ft/s.
Yes, because The driver is still above the speed limit after decelerating for 3 seconds.
To determine if you receive a ticket for speeding, we need to calculate the distance traveled during the deceleration period and compare it to the distance from the policeman.
First, convert the speed from miles per hour to feet per second. Since there are 5280 feet in a mile and 3600 seconds in an hour:
60 miles/hour = (60 * 5280) feet / (60 * 60) seconds = 88 feet/second.
Next, we calculate the initial velocity using the formula v = u + at, where:
v = final velocity,
u = initial velocity,
a = acceleration,
t = time.
Given that the initial velocity (u) is 88 feet/second, the acceleration (a) is -2 feet/second^2, and the time (t) is 3 seconds:
v = u + at
v = 88 + (-2 * 3)
v = 88 - 6
v = 82 feet/second.
Now, we can calculate the distance traveled during the deceleration period using the formula s = ut + (1/2)at^2, where:
s = distance,
u = initial velocity,
t = time,
a = acceleration.
Using the values u = 88 feet/second, t = 3 seconds, and a = -2 feet/second^2:
s = ut + (1/2)at^2
s = (88 * 3) + (1/2) * (-2) * (3^2)
s = 264 + (-2) * 9
s = 264 - 18
s = 246 feet.
So, you would have traveled 246 feet during the deceleration period.
If the policeman was only 10 feet ahead when you started decelerating, then you would have successfully slowed down and avoided getting a ticket since you would have stopped before reaching the policeman.