Question

An airplane needs to reach a velocity of 203.0 km/h to takeoff. On a 2000 m runway, what is the minimum accelerationnecessary for the plane to take flight? a) 0.79 m/s2 b) 1.0 m/s2 c) 0.95 m/s2 d) 0.87 m/s2

Measurement
ANSWERED
asked 2021-02-27
An airplane needs to reach a velocity of 203.0 km/h to takeoff. On a 2000 m runway, what is the minimum accelerationnecessary for the plane to take flight?
a) 0.79 m/s2
b) 1.0 m/s2
c) 0.95 m/s2
d) 0.87 m/s2

Expert Answers (2)

2021-02-28

We can solve using the equation
\(v_{f}^{2}= v_{0}^{2}+2a\triangle x\)
where ?x is the distance over shich the accelerationoccurs. Solving this for a, we have \(\displaystyle{a}={\frac{{{v}^{{{2}}}_{\left\lbrace{f}\right\rbrace}-{v}^{{{2}}}_{\left\lbrace{0}\right\rbrace}}}{{{2}\triangle{x}}}}\)
We need to convert our velocity to units of m/s, so first wemultiply by 1000 m/km and divide by 3600 s/h which gives
vf= \(\displaystyle{\frac{{{\left({203}{k}\frac{{m}}{{h}}\right)}{\left({1000}\frac{{m}}{{k}}{m}\right)}}}{{{\left({3600}\frac{{s}}{{h}}\right)}}}}={56.4}\frac{{m}}{{s}}\)
Now, we just insert \(v_{f}= 56.4\) m/s, \(v_{0}= 0\),and \(\triangle x = 2000\)m into our equation, giving \(\displaystyle{a}={\frac{{{\left({56.4}\frac{{m}}{{s}}\right)}^{{{2}}}}}{{{2}{\left({2000}{m}\right)}}}}={0.79}\frac{{m}}{{s}^{{{2}}}}\)
so our answer is a) 0.79 m/s

49
 
Best answer
2021-10-08

Step-by-step explanation:

Acceleration = change in Velocity over time. Mathematically it is represented as \(\frac{\triangle v}{ \triangle t}\)

Given values in the question:

  • Final velocity 203.0 km/h which is equal to (203/3.6)m/s = 56.39 m/s
  • distance 2000 m
  • Initial velocity is 0 m/s

To use the equation for acceleration time is needed.

velocity= distance/time

\((\frac{203}{3.6}) = \frac{2000}{time}\)  multiplying both sides by time

\((\frac{203}{3.6}) \times time = 2000\) making time subject of the formula

\(time = \frac{2000}{(\frac{203}{3.6})}\)

time = 35.47 seconds

Substituting into the acceleration equation

acceleration =\(\frac{\triangle v}{ \triangle t}\)

\(=\frac{(56.39- 0)}{35.47}\)

\(= \frac{56.39}{35.47}\)

\(=1.59 \ m/s^2\)

1

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