asked by @jeremyb10 •
7 months ago •
Physics
• 5 pts

Over a time interval of 2.06 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.4 km/s to -18.4 km/s. Find the following values, include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

A) Total change in the planet's velocity

B) Average acceleration during this interval

A) Total change=Vf-Vi=20.4-(-18.4)=-38.8 km/s

B)Average acceleration=Vf-Vi/t=38.8/t. t is time and is given in years therefore we should convert to seconds since velocity is given in units of seconds and convert km to m to get acceleration in SI units (m/s^2). 2.06 years=6.5x10^7 seconds. Therefore average acceleration=38.8x10^3/6.5x10^=-5.9x10^-4 m/s^2

answered by @juanv2 •
7 months ago

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