Minimum Deceleration for Railroad Crossing Accident Avoidance

AI Thread Summary
To determine the minimum deceleration required to avoid a railroad crossing accident, the engineer must first calculate the distance traveled during the reaction time of 0.22 seconds, which is 2.2 meters, leaving 207.8 meters to stop. The initial speed of the locomotive is 10 m/s, and the final speed must be 0 m/s to avoid collision. The relevant formula to find deceleration does not include time, allowing for a direct calculation based on distance and speed. The discussion emphasizes the importance of using the correct equations to solve for acceleration, highlighting common misconceptions about time in the calculations. Accurate application of physics principles is crucial for determining the necessary deceleration.
runner1738
Messages
71
Reaction score
0
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 teh car, the locomotive is 210 m from the crossing and its speed is 10 m/s. If the engineer's reaction time is .22 s, what should be the magnitude of the minmum deceleration to avoid an accident? Answer in units of m/s^2
 
Physics news on Phys.org
runner1738 said:
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 teh car, the locomotive is 210 m from the crossing and its speed is 10 m/s. If the engineer's reaction time is .22 s, what should be the magnitude of the minmum deceleration to avoid an accident? Answer in units of m/s^2
First, use the formula d=vt to find the distance he travels before applying the brakes. Subtract that distance from 210 m. Then, use the formula:
d=V_it + \frac{1}{2}at^2
to find the acceleration, which would be a negative number.
 
wait but shouldn't i solve for time 210=10(t) then t is 21 then 21-.22=20.78 then 210=10(20.78)+1/2(a)(20.78)^2 so a = .0101897345
 
No, since his speed is not 10 m/s throughout. First, find the DISTANCE he travels in 0.22s (d=(10 m/s)(0.22s)).
 
210-2.2=d=207.8=10(22)+1/2(a)(22)^2 ? ? or is 21.78 for time?
 
runner1738 said:
210-2.2=d=207.8=10(22)+1/2(a)(22)^2 ? ? or is 21.78 for time?
You got the first part (the 207.8m) right. As for the rest, you canot use 22s and the time, since you don't know the time it will take hime to stop. (Sorry, I gave you the wrong formula).
Ok, here's what you know:
initial velocity=10 m/s
final velocity = 0 m/s (since he has to stop)
d = 207.8 m
a=?

Can you find a formula that fits?
 
runner1738 said:
210-2.2=d=207.8=10(22)+1/2(a)(22)^2 ? ? or is 21.78 for time?

Where are you getting t=22?

I wouldn't use the above equation.

There's a different distance equation you can use and solve for "a" immediately... hint: it doesn't have a t in it.
 
how is time 17.21 seconds i have 207.8=10t-1/2(.2406)t^2
 

Similar threads

Finding deceleration of a moving train
Replies
7
Views
4K
Average Speed Question for a Train that Accelerates and then Decelerates
Replies
20
Views
2K
Help with question on motion: Avoiding a rear-end collision
Replies
6
Views
3K
Train collision prevention problem
Replies
8
Views
3K
History History of Railroad Safety - Spotlight on current derailments
2
Replies
58
Views
9K
How to find deceleration to stop collision with moving body?
Replies
6
Views
2K
Motion problem: motorist's braking time including reaction time
Replies
1
Views
2K
Deceleration of Train to Avoid Collision
Replies
1
Views
4K
What is the magnitude of the force?
Replies
1
Views
2K
Kinematics Problem: Train braking and blocking an intersection
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top