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

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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

Adnaan Franks · Accepted Answer

We can solve using the equation vf2=v02+2a△x where ?x is the distance over shich the accelerationoccurs. Solving this for a, we have a=v{f}2−v{0}22△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= (203kmh)(1000mkm)(3600sh)=56.4ms Now, we just insert vf=56.4 m/s, v0=0,and △x=2000m into our equation, giving a=(56.4ms)22(2000m)=0.79ms2 so our answer is a) 0.79 m/s

Jeffrey Jordon · Accepted Answer

Step-by-step explanation:  Acceleration = change in Velocity over time. Mathematically it is represented as △v△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  (2033.6)=2000time  multiplying both sides by time  (2033.6)×time=2000 making time subject of the formula  time=2000(2033.6)  time = 35.47 seconds  Substituting into the acceleration equation  acceleration =△v△t  =(56.39−0)35.47  =56.3935.47  =1.59 m/s2

user_27qwe · Accepted Answer

Step 1:First, we need to convert the velocities from km/h to m/s:v=203.0km/h=203.0×10003600m/s≈56.39m/su=0km/h=0m/sStep 2:Now, we can substitute these values into the equation:(56.39)2=(0)2+2×a×2000Simplifying the equation:3180.1521=4000aStep 3:Finally, solve for a:a=3180.15214000≈0.795m/s2Therefore, the minimum acceleration necessary for the plane to take flight is approximately 0.795m/s2.

karton · Accepted Answer

Answer:a) 0.79m/s2.Explanation:v2=u2+2asHere, v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.Given that the airplane needs to reach a velocity of 203.0 km/h (which can be converted to m/s) and the runway length is 2000 m, we can substitute these values into the equation.Let&#039;s calculate the minimum acceleration required for the plane to take flight.First, let&#039;s convert the velocity from km/h to m/s:203.0km/h=203.0×10003600m/sv=56.4m/sNext, we substitute the values into the equation:56.42=02+2×a×2000Simplifying the equation, we have:a=56.422×2000Calculating this value:a=56.424000a=0.7938m/s2Now, let&#039;s compare the calculated acceleration with the given options and find the closest one.a) 0.79m/s2b) 1.0m/s2c) 0.95m/s2d) 0.87m/s2The calculated minimum acceleration is approximately 0.7938m/s2, which is closest to option a) 0.79m/s2.

star233 · Accepted Answer

To solve the problem, we can use the following equation of motion:v2=u2+2aswhere:v = final velocity (203.0 km/h)u = initial velocity (0 km/h, as the plane starts from rest)a = acceleration (unknown)s = displacement (2000 m)We need to rearrange the equation to solve for a:a=v2−u22sSubstituting the given values:a=(203.0km/h)2−(0km/h)22·2000ma=(203.0km/h)24000ma=203.024000m/s2Calculating the value:a≈1.039m/s2Therefore, the minimum acceleration necessary for the plane to take flight is approximately 1.039m/s2.

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

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