Question

Over a time interval of 2.04 years, the velocity of a planet orbiting a distant star reverses direction, changing from +18.2 km/s to -22.9 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.

Answer 1

Explanation:

Initial speed of the planet, u = 18.2 km/s = 18200 m/s

Final speed of the planet, v = -22.9 km/s = -22900 m/s

Time interval, t = 2.04 years

Since, $1\ year =3.154\times 10^7\ s$

Time interval, $t=6.43\times 10^7\ s$

(a) The total change in the planet's velocity is calculated as :

$\Delta v=v-u$

$\Delta v=-22900-18200$

$\Delta v=-41100\ m/s$

(b) Let a is the average acceleration of the planet. It can be calculated as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{-41100\ m/s}{6.43\times 10^7\ s}$

$a=-6.39\times 10^{-4}\ m/s^2$

Hence, this is the required solution.