To determine the minimum deceleration needed to avoid an accident, we need to calculate the distance the train will travel before the engineer reacts and then finds the stopping distance needed to avoid hitting the car.
Step 1: Calculate the distance traveled during the engineer's reaction time.
The initial speed of the locomotive is \( v_0 = 12 , \text{m/s} \). The reaction time \( t_r = 0.32 , \text{s} \).
The distance traveled during the reaction time is given by:
\[ d_r = v_0 \cdot t_r = 12 , \text{m/s} \cdot 0.32 , \text{s} = 3.84 , \text{m} \]
Step 2: Calculate the remaining distance to the crossing after the reaction.
The total distance to the crossing is \( 290 , \text{m} \). After accounting for the distance traveled during the reaction time, the remaining distance \( d \) to the crossing is:
\[ d = 290 , \text{m} - 3.84 , \text{m} = 286.16 , \text{m} \]
Step 3: Determine the stopping distance required to come to a halt.
Let's assume the train decelerates at a rate of \( a , \text{m/s}^2 \). The stopping distance \( d_s \) from an initial velocity \( v_0 \) can be calculated using the equation:
\[ d_s = \frac{v_0^2}{2a} \]
We want the stopping distance to equal the remaining distance to the crossing, which means:
\[ \frac{v_0^2}{2a} = 286.16 , \text{m} \]
Substituting in \( v_0 = 12 , \text{m/s} \):
\[ \frac{(12 , \text{m/s})^2}{2a} = 286.16 , \text{m} \]
\[ \frac{144}{2a} = 286.16 \]
\[ \frac{72}{a} = 286.16 \]
Step 4: Solve for the deceleration \( a \).
Rearranging the equation gives:
\[ a = \frac{72}{286.16} \]
Calculating this value:
\[ a \approx 0.251 , \text{m/s}^2 \]
Conclusion:
The magnitude of the minimum deceleration needed to avoid an accident is approximately:
\[ a \approx 0.251 , \text{m/s}^2 \]
Thus, the final answer, rounded as appropriate, would be:
\[ \boxed{0.251} , \text{m/s}^2 \]