Question

A jet airplane touches down on an airport runway with a speed of 207 km/hr. It slows at a constant rate to a stop over a time period of 7.43 s. How far does the jet travel along the runway before stopping?

Answer 1

So distance travel by airplane before stopping is 213.637 m

Step-by-step explanation:

We have given initial speed of of the airplane u = 207 km /hr

Time t = 7.43 sec

207 km/hr =
`=(207* 1000m)/(3600sec)=57.5m/sec`

As the plane finally stops its final velocity v = 0 m/sec

According to first equation of motion v = u+at

So 0=57.5+a×7.43

a= -7.738
`m/sec^2`

Now according to third equation of motion
`v^2=u^2-2as`

`0^2=57.5^2-2* 7.738* s`

s = 213.637 m