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Sagot :
Sure! Let's break down the problem step-by-step:
1. Initial Information:
- Initial velocity [tex]\((v_i)\)[/tex]: [tex]\(72 \, \text{Hm/s}\)[/tex]
- Time to stop [tex]\((t)\)[/tex]: [tex]\(2 \, \text{seconds}\)[/tex]
- Mass of the truck [tex]\((m)\)[/tex]: [tex]\(5000 \, \text{kg}\)[/tex] (assuming the unit Hm is a typo).
2. Convert velocity from Hm/s to m/s:
- [tex]\(72 \, \text{Hm/s} = 72 \times 100 \, \text{m/s} = 7200 \, \text{m/s}\)[/tex]
3. Calculate the acceleration [tex]\((a)\)[/tex]:
- The truck comes to a stop, so the final velocity [tex]\((v_f)\)[/tex] is [tex]\(0 \, \text{m/s}\)[/tex].
- Using the formula [tex]\(v_f = v_i + at\)[/tex], rearrange to solve for [tex]\(a\)[/tex]:
[tex]\[ 0 = 7200 + a \times 2 \][/tex]
[tex]\[ a = -\frac{7200}{2} = -3600 \, \text{m/s}^2 \][/tex]
- The acceleration is negative because the truck is decelerating.
4. Calculate the distance covered while stopping:
- Using the kinematic equation:
[tex]\(\text{distance} = v_i \times t + \frac{1}{2} a \times t^2\)[/tex]
[tex]\[ \text{distance} = 7200 \times 2 + \frac{1}{2} \times (-3600) \times 2^2 \][/tex]
[tex]\[ \text{distance} = 14400 - 7200 = 7200 \, \text{m} \][/tex]
5. Calculate the force applied by the brakes:
- Using Newton's second law [tex]\(F = m \times a\)[/tex]:
[tex]\[ F = 5000 \times (-3600) \][/tex]
[tex]\[ F = -18000000 \, \text{N} \][/tex]
- The negative sign indicates that the force applied by the brakes is in the opposite direction of the truck's motion.
6. Summary of Results:
- Initial velocity in m/s: [tex]\(7200 \, \text{m/s}\)[/tex]
- Acceleration: [tex]\(-3600 \, \text{m/s}^2\)[/tex]
- Distance covered: [tex]\(7200 \, \text{m}\)[/tex]
- Force applied by the brakes: [tex]\(-18000000 \, \text{N}\)[/tex]
Thus, the values calculated are:
- The initial velocity is [tex]\(7200 \, \text{m/s}\)[/tex].
- The acceleration is [tex]\(-3600 \, \text{m/s}^2\)[/tex].
- The distance covered while stopping is [tex]\(7200 \, \text{m}\)[/tex].
- The force applied by the brakes is [tex]\(-18000000 \, \text{N}\)[/tex].
1. Initial Information:
- Initial velocity [tex]\((v_i)\)[/tex]: [tex]\(72 \, \text{Hm/s}\)[/tex]
- Time to stop [tex]\((t)\)[/tex]: [tex]\(2 \, \text{seconds}\)[/tex]
- Mass of the truck [tex]\((m)\)[/tex]: [tex]\(5000 \, \text{kg}\)[/tex] (assuming the unit Hm is a typo).
2. Convert velocity from Hm/s to m/s:
- [tex]\(72 \, \text{Hm/s} = 72 \times 100 \, \text{m/s} = 7200 \, \text{m/s}\)[/tex]
3. Calculate the acceleration [tex]\((a)\)[/tex]:
- The truck comes to a stop, so the final velocity [tex]\((v_f)\)[/tex] is [tex]\(0 \, \text{m/s}\)[/tex].
- Using the formula [tex]\(v_f = v_i + at\)[/tex], rearrange to solve for [tex]\(a\)[/tex]:
[tex]\[ 0 = 7200 + a \times 2 \][/tex]
[tex]\[ a = -\frac{7200}{2} = -3600 \, \text{m/s}^2 \][/tex]
- The acceleration is negative because the truck is decelerating.
4. Calculate the distance covered while stopping:
- Using the kinematic equation:
[tex]\(\text{distance} = v_i \times t + \frac{1}{2} a \times t^2\)[/tex]
[tex]\[ \text{distance} = 7200 \times 2 + \frac{1}{2} \times (-3600) \times 2^2 \][/tex]
[tex]\[ \text{distance} = 14400 - 7200 = 7200 \, \text{m} \][/tex]
5. Calculate the force applied by the brakes:
- Using Newton's second law [tex]\(F = m \times a\)[/tex]:
[tex]\[ F = 5000 \times (-3600) \][/tex]
[tex]\[ F = -18000000 \, \text{N} \][/tex]
- The negative sign indicates that the force applied by the brakes is in the opposite direction of the truck's motion.
6. Summary of Results:
- Initial velocity in m/s: [tex]\(7200 \, \text{m/s}\)[/tex]
- Acceleration: [tex]\(-3600 \, \text{m/s}^2\)[/tex]
- Distance covered: [tex]\(7200 \, \text{m}\)[/tex]
- Force applied by the brakes: [tex]\(-18000000 \, \text{N}\)[/tex]
Thus, the values calculated are:
- The initial velocity is [tex]\(7200 \, \text{m/s}\)[/tex].
- The acceleration is [tex]\(-3600 \, \text{m/s}^2\)[/tex].
- The distance covered while stopping is [tex]\(7200 \, \text{m}\)[/tex].
- The force applied by the brakes is [tex]\(-18000000 \, \text{N}\)[/tex].
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