Question

008 10.0 points 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 120 m from the crossing and its speed is 21 m/s. if the engineers reaction time is 0.64 s, what should be the magnitude of the minimum deceleration to avoid an accident? answer in units of m/s².

Explanation:

Step1: Calculate distance during reaction time

The train moves at constant speed during reaction time. Distance \( d_1 = v \times t_{reaction} \), where \( v = 21 \, \text{m/s} \), \( t_{reaction} = 0.64 \, \text{s} \).
\( d_1 = 21 \times 0.64 = 13.44 \, \text{m} \)

Step2: Calculate remaining distance to stop

Total distance to crossing is \( D = 120 \, \text{m} \). Remaining distance \( d_2 = D - d_1 = 120 - 13.44 = 106.56 \, \text{m} \)

Step3: Use kinematic equation to find deceleration

We use \( v_f^2 = v_i^2 + 2a d_2 \), where final velocity \( v_f = 0 \), initial velocity \( v_i = 21 \, \text{m/s} \), distance \( d_2 = 106.56 \, \text{m} \).
Rearranging for \( a \): \( a = \frac{v_f^2 - v_i^2}{2d_2} = \frac{0 - 21^2}{2 \times 106.56} \)
\( a = \frac{-441}{213.12} \approx -2.07 \, \text{m/s}^2 \) (magnitude is \( 2.07 \, \text{m/s}^2 \))

Answer:

\( 2.07 \) (or more precise value from calculation, e.g., if calculated as \( \frac{441}{213.12} \approx 2.07 \))