Question

A train is traveling along a straight, level track at $26.8 \mathrm{m} / \mathrm{s} \quad(60.0 \mathrm{mi} / \mathrm{h}) .$ Suddenly the engineer sees a truck stalled on the tracks \(184 \mathrm{m}\) ahead. If the maximum possible braking acceleration has magnitude \(1.52 \mathrm{m} / \mathrm{s}^{2},\) can the train be stopped in time?

Short answer

Answer: No, the train cannot stop in time before hitting the truck.

Step-by-step solution

Identifying the known variables

We are given the following information: Initial velocity (\(v_i\)) = \(26.8 \,\mathrm{m/s}\) Final velocity (\(v_f\)) = \(0 \,\mathrm{m/s}\) (since the train has to stop) Braking acceleration (\(a\)) = \(-1.52 \,\mathrm{m/s^2}\) (negative since it is decelerating) Distance to the truck (\(d_{truck}\)) = \(184 \,\mathrm{m}\)

Use the formula to find the stopping distance

To find the stopping distance (\(d_{stop}\)) of the train, we will use the following kinematic equation: \(v_f^2 = v_i^2 + 2ad_{stop}\)

Solve for the stopping distance (\(d_{stop}\))

We will now substitute the known variables into the equation and solve for the stopping distance (\(d_{stop}\)): \(0 = (26.8 \,\mathrm{m/s})^2 + 2 \times (-1.52 \,\mathrm{m/s^2}) d_{stop}\) Rearrange the equation to solve for \(d_{stop}\): \(d_{stop} = \frac{(26.8 \,\mathrm{m/s})^2}{2(1.52 \,\mathrm{m/s^2})}\) Calculate the value: \(d_{stop} \approx 236.3 \,\mathrm{m}\)

Compare stopping distance with distance to the truck

We now compare the stopping distance to the distance between the train and the truck: \(d_{stop} \approx 236.3 \,\mathrm{m}\) is greater than \(d_{truck} = 184 \,\mathrm{m}\) Since the stopping distance is greater than the distance to the truck, the train cannot stop in time before hitting the truck.