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

Step-by-step 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* 10^7\ s`

Time interval,
`t=6.43* 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=(v-u)/(t)`

`a=(-41100\ m/s)/(6.43* 10^7\ s)`

`a=-6.39* 10^(-4)\ m/s^2`

Hence, this is the required solution.