An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 290m from the crossing and it's speed is 20m/s . If the engineer's reaction time is 0.3s, What should be the magnitude of the minimum deceleration to avoid an accident?
2007-09-12 13:17:48 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
-0.7042 m/s^2.
After 0.3 s, the distance x = 290 - 20*0.3 = 284 m
Acceleration = -v^2/(2x) = -0.7042 m/s^2.
2007-09-12 13:28:36 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
v²=v0²-2ax where v is final speed, v0 is initial speed, a is accelleration and x is distance. Remember that x is -not- 290 m, it's 290-.3*20=284 m (to account for reaction time)
Now it's just plug 'n chug to find a such that V is 0.
FWIW, I got about .7 m/s²
Doug
2007-09-12 13:29:59 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
Zero. He can easily avoid an accident without slowing down at all by chosing to hit the car on purpose!
2007-09-12 14:15:10 · answer #3 · answered by Dr. R 7 · 0⤊ 0⤋
Vectors have 2 residences, value and direction. value is the quantity of rigidity, speed, or momentum. using 60 mph North or using 60 mph East has the comparable value, yet diverse direction. to describe the speed of an merchandise, you need to comprehend the fee and direction.
2016-11-15 02:07:24 · answer #4 · answered by ? 4 · 0⤊ 0⤋