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 110 m from the crossing and its speed is 16 m/s. if the engineer’s reaction time is 0.62 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 \( v = 16 \, \text{m/s} \), \( t_{reaction} = 0.62 \, \text{s} \).
\( d_1 = 16 \times 0.62 = 9.92 \, \text{m} \)

Step2: Calculate distance left to stop

Total distance to crossing is \( 110 \, \text{m} \). Distance left \( d_2 = 110 - d_1 = 110 - 9.92 = 100.08 \, \text{m} \)

Step3: Use kinematic equation to find deceleration

We use \( v_f^2 = v_i^2 + 2ad \), where \( v_f = 0 \) (final speed), \( v_i = 16 \, \text{m/s} \), \( d = d_2 = 100.08 \, \text{m} \).
Rearranging for \( a \): \( a = \frac{v_f^2 - v_i^2}{2d} = \frac{0 - 16^2}{2 \times 100.08} \)
\( a = \frac{-256}{200.16} \approx -1.279 \, \text{m/s}^2 \) (magnitude is positive, so \( 1.28 \, \text{m/s}^2 \) approximately)

Answer:

\( \approx 1.28 \, \text{m/s}^2 \) (or more precisely, using exact calculation: \( a = \frac{ - 256}{2\times(110 - 16\times0.62)} = \frac{ - 256}{2\times100.08} \approx -1.279 \), magnitude \( \approx 1.28 \))