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.

Question

Physics

Kayden Crane

1 January, 10:11

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.

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Answers (1)

Know the Answer?

Terrence Adkins · Answer 1

(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