The distance the jet traveled along the runway.

Question

Preparing for take-off, a jet starts at rest on the runway. The pilot fires up the engines, which accelerate the jet at a rate of 3.81 m/s². The jet lifts off the runway as it reaches a speed of 77.5 m/s. Find:

 (a) the time the jet was moving along the runway;

  (b) the distance the jet traveled along the runway.

Eqs:

0= v1 + at

x2= v1t + 1/2 at^2

Answer 1

To find the the time you can use the equation v_f=v_i+at where v_i=0 because jet starts from rest, v_f=77.5 m/s, and a=3.81 m/s². So, 77.5=0+3.81t, and solving for t you get: t=77.5/3.81=20.34 seconds.

To find distance you could use two equations. First one is: x=v_it+(1/2)at². Plug your t from previous part to find distance x=0+0.5(3.81)(20.34)²=788.13 meters.

You could also use the equation v_f²-v_i²=2ax without using the time. So, (77.5)²-0=2(3.81)x. Solving for x,

x=6006.25/7.62=788.22. Small decimal difference compared to previous equation is due to my rounding the numbers. But both give almost same answers.

Hope it helped.

Best,

Tomaj