Question

Over a time interval of 1.98 years, the velocity of a planet orbiting a distant star reverses direction, changing from +21.9 km/s to -17.2 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.

Please help!

Answer 1

change = -17.2 -21.9 = -39.1 *10^3 m/s

a = change in velocity/change in time

change in time
=1.98y(365d/y)(24h/d)(3600s/h)
=6.244128 *10^7 seconds
so
a = -3.91*10^4/6.24*10^7
= - 6.26*10^-4 m/s