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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 350 m from the crossing and its speed is 30 m/s. If the engineer's reaction time is 0.6 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s².

User Rad
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Final answer:

To avoid an accident, the locomotive needs to decelerate at a minimum magnitude of 1.22 m/s², taking into account the engineer's reaction time and the initial speed of the train.

Step-by-step explanation:

The question is asking for the magnitude of the minimum deceleration to avoid an accident, where an engineer in a locomotive sees a car stuck on the track at a railroad crossing. We must take into account both the reaction time of the engineer and the distance to the car to calculate the required deceleration. The formula to use is the equation of motion v^2 = u^2 + 2as, where v is the final velocity (0 m/s as the train must stop), u is the initial velocity (30 m/s), a is the deceleration, and s is the distance over which the train must stop. The distance must include the distance traveled during the engineer's reaction time. To find a, we rearrange the formula to a = (v^2 - u^2) / (2s). The train travels an additional u * reaction time during the engineer's reaction time. So, the total stopping distance is s + u * reaction time.

Let's calculate the stopping distance, including the reaction time:

s_total = s + u * reaction_time = 350 m + 30 m/s * 0.6 s = 350 m + 18 m = 368 m

Now substituting the values into the equation for a:

a = (0^2 - 30^2) / (2 * 368) = -900 / 736 = -1.22 m/s²

Therefore, the magnitude of minimum deceleration required to avoid an accident is 1.22 m/s².

User BIBEKRBARAL
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