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?

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

Where, v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the formula

We need to calculate the distance from the original runway

Using equation of motion

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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)