# Magnitude of deceleration

1. Feb 4, 2005

### runner1738

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

2. Feb 4, 2005

### christinono

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.

3. Feb 4, 2005

### runner1738

wait but shouldnt 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

4. Feb 4, 2005

### christinono

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)).

5. Feb 4, 2005

### runner1738

210-2.2=d=207.8=10(22)+1/2(a)(22)^2 ? ? or is 21.78 for time?

6. Feb 4, 2005

### christinono

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?

7. Feb 4, 2005

### learningphysics

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.

8. Feb 5, 2005

### runner1738

how is time 17.21 seconds i have 207.8=10t-1/2(.2406)t^2