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Sagot :
Certainly! Let's break down the problem step-by-step:
1. Initial Acceleration Phase:
- The particle starts from rest, meaning its initial velocity ([tex]\(u\)[/tex]) is 0.
- It accelerates uniformly at [tex]\(6 \, \text{m/s}^2\)[/tex] for [tex]\(8 \, \text{s}\)[/tex].
We need to find the speed at the end of this acceleration phase. We can use the equation of motion:
[tex]\[ v = u + at \][/tex]
where:
- [tex]\(v\)[/tex] is the final velocity
- [tex]\(u\)[/tex] is the initial velocity (0 in this case)
- [tex]\(a\)[/tex] is the acceleration ([tex]\(6 \, \text{m/s}^2\)[/tex])
- [tex]\(t\)[/tex] is the time ([tex]\(8 \, \text{s}\)[/tex])
Substituting the given values:
[tex]\[ v = 0 + (6 \, \text{m/s}^2 \times 8 \, \text{s}) = 48 \, \text{m/s} \][/tex]
Therefore, the speed at the end of the acceleration phase is [tex]\(48 \, \text{m/s}\)[/tex].
2. Deceleration Phase:
- The particle then decelerates uniformly and comes to rest in the next [tex]\(5 \, \text{s}\)[/tex].
- The initial velocity for this phase is the final velocity from the acceleration phase, which is [tex]\(48 \, \text{m/s}\)[/tex].
- The final velocity ([tex]\(v\)[/tex]) at the end of deceleration is [tex]\(0 \, \text{m/s}\)[/tex].
We need to calculate the deceleration ([tex]\(a\)[/tex]).
Using the equation of motion again:
[tex]\[ v = u + at \][/tex]
Given:
- [tex]\(v\)[/tex] is the final velocity (0 \, \text{m/s})
- [tex]\(u\)[/tex] is the initial velocity ([tex]\(48 \, \text{m/s}\)[/tex])
- [tex]\(t\)[/tex] is the time ([tex]\(5 \, \text{s}\)[/tex])
Rearranging the equation to solve for [tex]\(a\)[/tex]:
[tex]\[ 0 = 48 + a \times 5 \][/tex]
[tex]\[ a \times 5 = -48 \][/tex]
[tex]\[ a = - \frac{48}{5} = -9.6 \, \text{m/s}^2 \][/tex]
The negative sign indicates that it is a deceleration.
Thus, the speed after the acceleration phase is [tex]\(48 \, \text{m/s}\)[/tex], and since the particle comes to rest, this also means that it decelerates uniformly over the next [tex]\(5 \, \text{s}\)[/tex].
So, to answer the multiple-choice question:
- The correct answer is B. [tex]\(48 \, \text{m/s}\)[/tex].
1. Initial Acceleration Phase:
- The particle starts from rest, meaning its initial velocity ([tex]\(u\)[/tex]) is 0.
- It accelerates uniformly at [tex]\(6 \, \text{m/s}^2\)[/tex] for [tex]\(8 \, \text{s}\)[/tex].
We need to find the speed at the end of this acceleration phase. We can use the equation of motion:
[tex]\[ v = u + at \][/tex]
where:
- [tex]\(v\)[/tex] is the final velocity
- [tex]\(u\)[/tex] is the initial velocity (0 in this case)
- [tex]\(a\)[/tex] is the acceleration ([tex]\(6 \, \text{m/s}^2\)[/tex])
- [tex]\(t\)[/tex] is the time ([tex]\(8 \, \text{s}\)[/tex])
Substituting the given values:
[tex]\[ v = 0 + (6 \, \text{m/s}^2 \times 8 \, \text{s}) = 48 \, \text{m/s} \][/tex]
Therefore, the speed at the end of the acceleration phase is [tex]\(48 \, \text{m/s}\)[/tex].
2. Deceleration Phase:
- The particle then decelerates uniformly and comes to rest in the next [tex]\(5 \, \text{s}\)[/tex].
- The initial velocity for this phase is the final velocity from the acceleration phase, which is [tex]\(48 \, \text{m/s}\)[/tex].
- The final velocity ([tex]\(v\)[/tex]) at the end of deceleration is [tex]\(0 \, \text{m/s}\)[/tex].
We need to calculate the deceleration ([tex]\(a\)[/tex]).
Using the equation of motion again:
[tex]\[ v = u + at \][/tex]
Given:
- [tex]\(v\)[/tex] is the final velocity (0 \, \text{m/s})
- [tex]\(u\)[/tex] is the initial velocity ([tex]\(48 \, \text{m/s}\)[/tex])
- [tex]\(t\)[/tex] is the time ([tex]\(5 \, \text{s}\)[/tex])
Rearranging the equation to solve for [tex]\(a\)[/tex]:
[tex]\[ 0 = 48 + a \times 5 \][/tex]
[tex]\[ a \times 5 = -48 \][/tex]
[tex]\[ a = - \frac{48}{5} = -9.6 \, \text{m/s}^2 \][/tex]
The negative sign indicates that it is a deceleration.
Thus, the speed after the acceleration phase is [tex]\(48 \, \text{m/s}\)[/tex], and since the particle comes to rest, this also means that it decelerates uniformly over the next [tex]\(5 \, \text{s}\)[/tex].
So, to answer the multiple-choice question:
- The correct answer is B. [tex]\(48 \, \text{m/s}\)[/tex].
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