a plane lands at a speed of 68 m/s and slows down at a rate 0f 4 m/s^2. How much runway is needed to stop the plane?
Answer
By applying third equation of motion
"v^2=u^2+2as"
When plane will stop then
"v=0m\/s"
Acceleration
"a=-4m\/s^2"
Initial velocity
"u=68m\/s"
So by putting value
"0^2=68^2-2(4) (s) \\\\s=\\frac{68^2}{8}\\\\s=578m"
So 578m runway is required.
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