At the starting gun a runner accelerates from rest at 1.7 m/s^2 for 2.2 what is the runner's speed 1.3 s after she starts running?

The runner´s speed is 2.2 m/s after 1.3 s.

Explanation:

The acceleration is the variation of velocity over time. If the acceleration is constant, then:

dv/dt = a

Separating variables:

dv = a · dt

integrating from initial velocity (v0) to v and from t = 0 to t:

v - v0 = a · t

v = v0 + a · t

Then usign this equation, we can calcualte the speed at t = 1.3 s:

v = 0 m/s + 1.7 m/s² · 1.3 s

v = 2.2 m/s

The runner´s speed is 2.2 m/s after 1.3 s.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

the part of the wave he is travelling on is the crest