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Sagot :
To find the displacement of the particle, we can use the kinematic equation for uniformly accelerated motion. The equation is:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the displacement,
- [tex]\( u \)[/tex] is the initial velocity,
- [tex]\( a \)[/tex] is the constant acceleration,
- [tex]\( t \)[/tex] is the time.
Now let's fill in the given values:
- The initial velocity ([tex]\( u \)[/tex]) is [tex]\( 5 \, \text{m/s} \)[/tex].
- The constant acceleration ([tex]\( a \)[/tex]) is [tex]\( 0.2 \, \text{m/s}^2 \)[/tex].
- The time ([tex]\( t \)[/tex]) is [tex]\( 15 \, \text{seconds} \)[/tex].
We plug in these values into the formula:
1. Calculating [tex]\( ut \)[/tex]:
[tex]\[ ut = 5 \, \text{m/s} \times 15 \, \text{seconds} = 75 \, \text{meters} \][/tex]
2. Calculating [tex]\( \frac{1}{2}at^2 \)[/tex]:
[tex]\[ \frac{1}{2} \times 0.2 \, \text{m/s}^2 \times (15 \, \text{seconds})^2 \][/tex]
[tex]\[ \frac{1}{2} \times 0.2 \times 225 \][/tex]
[tex]\[ \frac{1}{2} \times 45 \][/tex]
[tex]\[ 22.5 \, \text{meters} \][/tex]
3. Adding both terms to get the total displacement [tex]\( s \)[/tex]:
[tex]\[ s = 75 \, \text{meters} + 22.5 \, \text{meters} \][/tex]
[tex]\[ s = 97.5 \, \text{meters} \][/tex]
Therefore, the displacement of the particle in 15 seconds is [tex]\( 97.5 \)[/tex] meters.
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the displacement,
- [tex]\( u \)[/tex] is the initial velocity,
- [tex]\( a \)[/tex] is the constant acceleration,
- [tex]\( t \)[/tex] is the time.
Now let's fill in the given values:
- The initial velocity ([tex]\( u \)[/tex]) is [tex]\( 5 \, \text{m/s} \)[/tex].
- The constant acceleration ([tex]\( a \)[/tex]) is [tex]\( 0.2 \, \text{m/s}^2 \)[/tex].
- The time ([tex]\( t \)[/tex]) is [tex]\( 15 \, \text{seconds} \)[/tex].
We plug in these values into the formula:
1. Calculating [tex]\( ut \)[/tex]:
[tex]\[ ut = 5 \, \text{m/s} \times 15 \, \text{seconds} = 75 \, \text{meters} \][/tex]
2. Calculating [tex]\( \frac{1}{2}at^2 \)[/tex]:
[tex]\[ \frac{1}{2} \times 0.2 \, \text{m/s}^2 \times (15 \, \text{seconds})^2 \][/tex]
[tex]\[ \frac{1}{2} \times 0.2 \times 225 \][/tex]
[tex]\[ \frac{1}{2} \times 45 \][/tex]
[tex]\[ 22.5 \, \text{meters} \][/tex]
3. Adding both terms to get the total displacement [tex]\( s \)[/tex]:
[tex]\[ s = 75 \, \text{meters} + 22.5 \, \text{meters} \][/tex]
[tex]\[ s = 97.5 \, \text{meters} \][/tex]
Therefore, the displacement of the particle in 15 seconds is [tex]\( 97.5 \)[/tex] meters.
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