Question

The captain orders his starship to accelerate from rest at a rate of "1 g" (1 g = 9.8 m/s2). How many days does it take the starship to reach 10% the speed of light? (Light travels at 3.0 × 108 m/s.)

Answer 1

Answer:35days

Explanation: acceleration a=velocity v/time

a is. given to be 9.8m/s2

V=0.1x speed of light,=3.0x107

t =3.0x10*7/9.8

= 3.06x10*6sec

converting to days I. E divide by 86400 will make t to be 35.days to the nearest significant number.

Answer 2

Answer:

The  days it will take the starship to reach 10% of speed of light is approximately 35 days.

Explanation:

Given;

acceleration of the starship, a = 9.8 m/s²

speed of light is given as, v = 3.0 × 10⁸ m/s

10% of speed light = 0.1 x 3.0 × 10⁸ m/s = 3.0 × 10⁷ m/s

Apply kinematic equation;

acceleration is the time rate of change in velocity.

a = Δv/t

t = Δv/a

Substitute the given values and solve for time

t =  (3.0 × 10⁷)/9.8

t = 3061224.4898 seconds

$time \ in \ days = 3061224.4898 (s)*\frac{1hr}{3600s}*\frac{1 day}{24hrs} = 35.43 \ days$

Therefore, the  days it will take the starship to reach 10% of speed of light is approximately 35 days.