Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/s to -22.0 km/s. find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

                     = ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

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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.