Question

g Over a time interval of 2.28 years, the velocity of a planet orbiting a distant star reverses direction, changing from +22.4 km/s to -17.5 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

(A) ΔV = –39900m/s

(B) a = –5.55 ×10-⁴ m/s²

Step-by-step explanation:

ΔV = V2 - V1 = –17.5 –22.4 = ‐39.9km/s = –39.9 ×1000m/s

= –39900m/s

Δt = 2.28years = 2.28 ×365.25days

= 2.28 × 365.25 × 86400s = 71951328s

a = ΔV/Δt = ‐39900m/71951328s = –5.55 × 10-⁴m/s²