Question

After touchdown a fighter jet passes a marker on the runway at an instantaneous speed of 100 m/s and constant negative acceleration of -15 m/s2. After slowing down to 25 m/s, the pilot suddenly goes to full power to provide a new constant acceleration of 35 m/s2 for a "touch and go" takeoff. How far from the original runway marker does the plane achieve its takeoff speed of 170 m/s?

Answer 1

Given that,

Deacelectration = -15 m/s²

Negative sign shows the declaration

Slowly speed = 25 m/s

Acceleration = 35 m/s²

Speed = 170 m/s

We need to calculate the time

Using equation of motion

$v=u+at$

Where, v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the formula

$25=0+15\times t$

$t=\dfrac{25}{15}$

$t=1.66\ sec$

We need to calculate the distance from the original runway

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times35\times(1.66)^2$

$s=48.2\ m$

Hence, The distance from the original runway is 48.2 m.