Question

A space ship zips past Mars’ orbit at a speed of 3 x 103 m/s and constantly accelerates in a straight line until it reaches the orbital path of Jupiter, 5.56 x 1011 m away, when its speed is 4 x 104 m/s. Find A) the space ship’s acceleration and B) the time required to travel from Mars’ to Jupiter’s orbits. (Answer in days

Answer 1

Step-by-step explanation:

given,

space ship speed = 3 x 10³ m/s

distance travel in straight line = 5.56 x 10¹¹ m

speed at that point = 4 x 10⁴ m/s

a) acceleration

v² - u² = 2 a s

(4 x 10⁴)² - (3 x 10³)² = 2 x a x 5.56 x 10¹¹

`a = (0.01591* 10^(11))/(11.12 * 10^(11))`

a = 1.43 x 10⁻³ m/s²

b) time require to travel mars

v + u = a t

4 x 10⁴ + 3 x 10³ = (1.43 x 10⁻³) t

t = 30069930 sec

`t = (30069930\ sec)/(86400)\ days`

t = 348 days