An unmarked police car, traveling a constant 87.0 km/hr, is passed by a speeder traveling 151 km/hr. Precisely 1.44 s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.08 m/s2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)?
How would you solve this problem?
2006-10-19 08:19:00 · 4 answers · asked by beautyqueenjustine 3 in Science & Mathematics ➔ Physics
At the 1.44s point, the speeder is (151000 - 81000)*1.44/3600 = 28m ahead and the relative speed is 70000/3600 = 19.444 m/s.
The relative distance betwen cars is therefore
Dr = 28 + 19.444t -(2.08/2)t^2
The desired distance is zero, so 1.04t^2 - 19.444t - 28 = 0
Use the quadratic formula to solve for t.
I got t = 20.84 sec or -1.397 sec
Add 1.44 sec to the 20.84 to answer the ? just the way you asked it!
2006-10-19 10:16:08 · answer #1 · answered by Steve 7 · 0⤊ 0⤋
Why worry yourself or be concerned of a possible speed trap if you are traveling at the speed the law permits? Speeding will only land you in one place - the morgue Maintain a safe speed keeping a safe distance between cars and stop often for breaks. Have a safe day, Don't know why Joe or anyother person would tell you otherwise.
2016-03-28 01:45:00 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Distance traveled for speeder: vs*t
Distance traveled for police vp*t+(1/2)a*(t-1.44s)^2
2006-10-19 08:28:47 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋
hmmmmmmmmmm i know the answer:
one big high speed chase
2006-10-19 08:23:55 · answer #4 · answered by craftyboy 2 · 0⤊ 0⤋