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 40 m/s2 for a "touch and go" takeoff. how far from the original runway marker does the plane achieve its takeoff speed of 162 m/s?

632 m.

Explanation:

We can divide the problem (and hence the total displacement over the runway) in two parts:

1) The information that we have, is the initial velocity, the acceleration value, and the final value for the velocity during this part.

So, applying acceleration definition, we can find the time used in order to travel this part, as follows:

a = (vf - v₀) / t ⇒ t = (vf-v₀) / a = (25 m/s -100 m/s) / -15 m/s² = 5 s

Now, we can replace this value in the kinematic equation for displacement:

x = v₀ . t + 1/2 a t²

Replacing by the values, we get:

x= 100 m/s. 5s + 1/2 (-15 m/s²). (5²) s² = 500 m - 188 m = 312 m.

2) For this part, we have the values of acceleration, and of the final and initial velocity values, so we can find again the time for this part:

t = 3.43 s

Replacing in the equation for displacement, we have the distance traveled in this second part:

x = 25 m/s. (3.43s) + 1/2 (40 m/s²). (3.43)² s² = 320 m.

Explanation:

b. latent learning (the word "latent" itself means something that exists but isn't obvious, only under certain conditions! )

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat