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 thecar, the locomotive is 310 m from the crossingand its speed is 10 m/s.If the engineer’s reaction time is 0.39 s,what should be the magnitude of the minimum deceleration to avoid an accident?Answer in units of m/s2.
Well, if the engineer wants to avoid a collision with the car stuck on the track, he better hit the brakes! To calculate the minimum deceleration needed, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Here, the final velocity would be zero (as the train needs to come to a stop), the initial velocity is 10 m/s (given in the problem), and the time is the reaction time of 0.39 seconds. Plugging in the values:
acceleration = (0 - 10) m/s / 0.39 s
Solving this equation, the minimum deceleration required to avoid an accident is approximately -25.64 m/s^2. But don't worry, if the engineer is skilled, he can handle this emergency situation like a pro!
To calculate the minimum deceleration needed to avoid an accident, we can use the formula:
\(d = \frac{{v_f^2 - v_i^2}}{{2a}}\)
where:
d = distance traveled
v_f = final velocity
v_i = initial velocity
a = acceleration (deceleration in this case)
Given:
\(v_i = 10 \, \text{m/s}\),
\(d = 310 \, \text{m}\),
\(t = 0.39 \, \text{s}\)
First, let's calculate the final velocity using the formula:
\(v_f = v_i + at\)
\(v_f = 0 + 10a\)
Now we can substitute the values into the first formula and solve for the acceleration:
\(d = \frac{{(10a)^2 - 10^2}}{{2a}}\)
\(d = \frac{{100a^2 - 100}}{{2a}}\)
\(d = \frac{{100a(a - 1)}}{{2a}}\)
Next, we can cancel out the common factor of \(a\) and solve for \(a\):
\(d = 50(a - 1)\)
\(a - 1 = \frac{d}{50}\)
\(a = \frac{d}{50} + 1\)
Substituting the value of \(d = 310\, \text{m}\):
\(a = \frac{310}{50} + 1\)
\(a = 6.2 + 1\)
\(a = 7.2 \, \text{m/s}^2\)
Therefore, the magnitude of the minimum deceleration required to avoid an accident is 7.2 m/s².
To find the magnitude of the minimum deceleration required to avoid an accident, we can follow these steps:
1. Convert the given distance of 310 m to a negative value since it is in the opposite direction of the locomotive's motion.
- Distance to car = -310 m
2. Determine the total distance covered by the locomotive during the reaction time.
- Distance covered during reaction time = speed × reaction time
- Distance covered during reaction time = 10 m/s × 0.39 s
3. Calculate the remaining distance between the locomotive and the car after the reaction time.
- Remaining distance = distance to car - distance covered during reaction time
- Remaining distance = -310 m - (10 m/s × 0.39 s)
4. Calculate the minimum deceleration required to stop the train within the remaining distance.
- Final velocity of the locomotive = 0 m/s (since it needs to come to a stop)
- Using the kinematic equation: v² = u² + 2as
- where v = final velocity (0 m/s), u = initial velocity (10 m/s), a = acceleration/deceleration, and s = distance
- Rearranging the equation, we get: a = (v² - u²) / (2s)
- Minimum deceleration = (0 m/s - (10 m/s)²) / (2 × remaining distance)
5. Substitute the values into the equation to find the minimum deceleration.
- Minimum deceleration = (0 - 100) / (2 × remaining distance)
Now, let's substitute the values and calculate the minimum deceleration:
Remaining distance = -310 m - (10 m/s × 0.39 s)
Remaining distance = -310 m - 3.9 m
Remaining distance = -313.9 m
Minimum deceleration = (0 - 100) / (2 × remaining distance)
Minimum deceleration = -100 / (2 × (-313.9 m))
Minimum deceleration ≈ -0.159 m/s²
Since the question asks for the magnitude of the deceleration, we take the absolute value:
Magnitude of the minimum deceleration ≈ 0.159 m/s²
Therefore, the magnitude of the minimum deceleration required to avoid an accident is approximately 0.159 m/s².
D = 310 m -10m/s *0.39s = 306.1 m
after reaction time.
D = v * t
306.1 = 10m/s * t
t = 30.61s = time to reach stalled car.
Deceleration must reduce velocity to
0 in 30.6 s or less.
Deceleration = (Vt - Vo) / t =
(0 - 10m/s) / 30.61s = -0.327m/s^2.