Answered

a) [tex]\(12.5 \, \text{km}\)[/tex]

b) [tex]\(91.4 \, \text{km}\)[/tex]

c) [tex]\(11.3 \, \text{km}\)[/tex]

d) [tex]\(49.2 \, \text{km}\)[/tex]

Given:
[tex]\[ d = V \cdot t \][/tex]
[tex]\[ 45 \, \text{m} \cdot \frac{60 \, \text{s}}{1 \, \text{m}} - 2 \][/tex]
[tex]\[ 15 \, \text{km/h} = 11.30 \][/tex]

5. A train travels at [tex]\(60 \, \text{km/h}\)[/tex]. Immediately after passing a red signal, it begins to come to a complete stop over a distance of 150 meters. Determine its acceleration.

a) [tex]\(-0.5 \, \text{m/s}^2\)[/tex]



Answer :

GinnyAnswer
Sure, let's solve the problem.

Question:
A train travels at a speed of 60 km/h. Immediately after passing a red signal, it starts to come to a complete stop within a distance of 150 meters. Determine its acceleration.

Solution:

1. Initial Information:
- Initial velocity ([tex]\(u\)[/tex]) = 60 km/h
- Stopping distance ([tex]\(s\)[/tex]) = 150 meters

2. Convert the initial velocity from km/h to m/s:
- To convert km/h to m/s, you can use the formula:
[tex]\[ \text{Velocity in m/s} = \left( \frac{\text{Velocity in km/h} \times 1000}{3600} \right) \][/tex]
- Plugging in the values:
[tex]\[ u = \frac{60 \times 1000}{3600} = 16.67 \text{ m/s} \][/tex]

3. Use the kinematic equation to find the acceleration:
- The kinematic equation which relates initial velocity, final velocity, acceleration, and distance is:
[tex]\[ v^2 = u^2 + 2as \][/tex]
- Here, final velocity ([tex]\(v\)[/tex]) is 0 because the train comes to a stop.
- Rearrange to solve for acceleration ([tex]\(a\)[/tex]):
[tex]\[ 0 = u^2 + 2as \implies a = -\frac{u^2}{2s} \][/tex]
- Substitute the known values into the equation:
[tex]\[ a = -\frac{(16.67)^2}{2 \times 150} \approx -0.93 \text{ m/s}^2 \][/tex]

Answer:
The acceleration of the train is approximately [tex]\(-0.93 \text{ m/s}^2\)[/tex].

This solution accounts for the initial velocity conversion and the application of the kinematic equation to determine the train's deceleration as it stops after passing the red signal.

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