Free Fall/Constant Acceleration Problem

[cheerspens]

Homework Statement

A rocket ship in free space travels with a constant acceleration equal to 9.8m/s^s.
a)If it starts from rest how long does it take to reach a velocity of one tenth of the speed of light? Note: The speed of light in a vacuum is 3.0 x 10^8 m/s.
b)How far will it travel during this time?

Homework Equations

V=Vo+at
X=Xo+Vot+(1/2)(a)(t)

The Attempt at a Solution

I started with making a variable list:
a=9.8 t=?
Vo=0 Xo=0
V=30,000,000 X=?

I plugged these values into the equation I mentioned in # 2.
The answers I got are:
a) 3061224.49sec
b)4.59 x 10^13
I wasn't sure how to do this and gave it a shot. Could you please explain.
Thanks!

 

[Doc Al]

Your method is fine, but be careful with decimal points. Your answer for (a) is off by a factor of 10, making (b) off by a factor of 100.

 

[kerimek]

I would like to commend you for your attempt at solving this problem! You have correctly used the equations for constant acceleration to find the time and distance traveled by the rocket ship.

However, there are a few things to consider when solving problems involving free fall or constant acceleration. First, it is important to ensure that the units used for all variables are consistent. In this case, the acceleration is given in meters per second squared (m/s^2) and the velocity is given in meters per second (m/s). Therefore, the time should be in seconds (s) and the distance should be in meters (m).

Secondly, since the speed of light is a very large value, it is important to use scientific notation to avoid errors. In this case, the speed of light should be written as 3.0 x 10^8 m/s, not 30000000 m/s.

With these corrections, the equations become:

V = Vo + at
X = Xo + Vot + (1/2)at^2

Now, let's solve for the time it takes for the rocket to reach one tenth of the speed of light (3.0 x 10^7 m/s):

V = 3.0 x 10^7 m/s
Vo = 0 m/s
a = 9.8 m/s^2

Using the first equation, we can solve for time:

3.0 x 10^7 m/s = 0 m/s + (9.8 m/s^2)t

t = (3.0 x 10^7 m/s) / (9.8 m/s^2) = 3.06 x 10^6 s

This is equivalent to 3.06 million seconds, or approximately 35.4 days. So, it would take the rocket about 35 days to reach one tenth of the speed of light with a constant acceleration of 9.8 m/s^2.

To find the distance traveled during this time, we can use the second equation:

X = 0 m + (3.0 x 10^7 m/s)(3.06 x 10^6 s) + (1/2)(9.8 m/s^2)(3.06 x 10^6 s)^2

X = 4.59 x 10^13 m

This is equivalent to approximately 4.