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Sagot :
Sure, let's solve this problem step by step.
### Given Data:
- Initial velocity ([tex]\(u\)[/tex]) of the aeroplane: [tex]\(216 \, \text{km/h}\)[/tex]
- Distance ([tex]\(s\)[/tex]) the aeroplane covers before coming to rest: [tex]\(2 \, \text{km}\)[/tex]
- Final velocity ([tex]\(v\)[/tex]) of the aeroplane when it comes to rest: [tex]\(0 \, \text{m/s}\)[/tex]
### Convert the Initial Velocity to [tex]\(m/s\)[/tex]:
Since velocity is given in [tex]\( \text{km/h} \)[/tex], we need to convert it to [tex]\( \text{m/s} \)[/tex]:
[tex]\[ u = 216 \, \text{km/h} = 216 \times \left(\frac{1000 \, \text{m}}{1 \, \text{km}}\right) \times \left(\frac{1 \, \text{h}}{3600 \, \text{s}}\right) \][/tex]
[tex]\[ u = 216 \times \left(\frac{1000}{3600}\right) \, \text{m/s} \][/tex]
[tex]\[ u = 216 \times \left(\frac{5}{18}\right) \, \text{m/s} \][/tex]
[tex]\[ u = 216 \div 3.6 \, \text{m/s} \][/tex]
[tex]\[ u = 60 \, \text{m/s} \][/tex]
### Convert Distance to Meters:
Since distance is given in [tex]\( \text{km} \)[/tex], we need to convert it to [tex]\( \text{m} \)[/tex]:
[tex]\[ s = 2 \, \text{km} = 2 \times 1000 \, \text{m} \][/tex]
[tex]\[ s = 2000 \, \text{m} \][/tex]
### Calculate the Acceleration:
We use the equation of motion:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Given that the final velocity [tex]\(v = 0 \, \text{m/s}\)[/tex]:
[tex]\[ 0 = (60)^2 + 2a \times 2000 \][/tex]
[tex]\[ 0 = 3600 + 4000a \][/tex]
[tex]\[ 4000a = -3600 \][/tex]
[tex]\[ a = \frac{-3600}{4000} \][/tex]
[tex]\[ a = -0.9 \, \text{m/s}^2 \][/tex]
### Calculate the Time to Come to Rest:
We use another equation of motion:
[tex]\[ v = u + at \][/tex]
Given that the final velocity [tex]\(v = 0 \, \text{m/s}\)[/tex]:
[tex]\[ 0 = 60 + (-0.9)t \][/tex]
[tex]\[ -60 = -0.9t \][/tex]
[tex]\[ t = \frac{60}{0.9} \][/tex]
[tex]\[ t = 66.67 \, \text{s} \][/tex]
### Final Answers:
- Acceleration [tex]\(a\)[/tex] is:
[tex]\[ a = -0.9 \, \text{m/s}^2 \][/tex]
- Time [tex]\(t\)[/tex] to come to rest is:
[tex]\[ t = 66.67 \, \text{s} \][/tex]
So, the acceleration of the aeroplane is [tex]\(-0.9 \, \text{m/s}^2\)[/tex], and the time taken to come to rest is [tex]\(66.67\)[/tex] seconds.
### Given Data:
- Initial velocity ([tex]\(u\)[/tex]) of the aeroplane: [tex]\(216 \, \text{km/h}\)[/tex]
- Distance ([tex]\(s\)[/tex]) the aeroplane covers before coming to rest: [tex]\(2 \, \text{km}\)[/tex]
- Final velocity ([tex]\(v\)[/tex]) of the aeroplane when it comes to rest: [tex]\(0 \, \text{m/s}\)[/tex]
### Convert the Initial Velocity to [tex]\(m/s\)[/tex]:
Since velocity is given in [tex]\( \text{km/h} \)[/tex], we need to convert it to [tex]\( \text{m/s} \)[/tex]:
[tex]\[ u = 216 \, \text{km/h} = 216 \times \left(\frac{1000 \, \text{m}}{1 \, \text{km}}\right) \times \left(\frac{1 \, \text{h}}{3600 \, \text{s}}\right) \][/tex]
[tex]\[ u = 216 \times \left(\frac{1000}{3600}\right) \, \text{m/s} \][/tex]
[tex]\[ u = 216 \times \left(\frac{5}{18}\right) \, \text{m/s} \][/tex]
[tex]\[ u = 216 \div 3.6 \, \text{m/s} \][/tex]
[tex]\[ u = 60 \, \text{m/s} \][/tex]
### Convert Distance to Meters:
Since distance is given in [tex]\( \text{km} \)[/tex], we need to convert it to [tex]\( \text{m} \)[/tex]:
[tex]\[ s = 2 \, \text{km} = 2 \times 1000 \, \text{m} \][/tex]
[tex]\[ s = 2000 \, \text{m} \][/tex]
### Calculate the Acceleration:
We use the equation of motion:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Given that the final velocity [tex]\(v = 0 \, \text{m/s}\)[/tex]:
[tex]\[ 0 = (60)^2 + 2a \times 2000 \][/tex]
[tex]\[ 0 = 3600 + 4000a \][/tex]
[tex]\[ 4000a = -3600 \][/tex]
[tex]\[ a = \frac{-3600}{4000} \][/tex]
[tex]\[ a = -0.9 \, \text{m/s}^2 \][/tex]
### Calculate the Time to Come to Rest:
We use another equation of motion:
[tex]\[ v = u + at \][/tex]
Given that the final velocity [tex]\(v = 0 \, \text{m/s}\)[/tex]:
[tex]\[ 0 = 60 + (-0.9)t \][/tex]
[tex]\[ -60 = -0.9t \][/tex]
[tex]\[ t = \frac{60}{0.9} \][/tex]
[tex]\[ t = 66.67 \, \text{s} \][/tex]
### Final Answers:
- Acceleration [tex]\(a\)[/tex] is:
[tex]\[ a = -0.9 \, \text{m/s}^2 \][/tex]
- Time [tex]\(t\)[/tex] to come to rest is:
[tex]\[ t = 66.67 \, \text{s} \][/tex]
So, the acceleration of the aeroplane is [tex]\(-0.9 \, \text{m/s}^2\)[/tex], and the time taken to come to rest is [tex]\(66.67\)[/tex] seconds.
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