# acceleration, 0 to 669,600,000 MPH at 1G?

i would like to know the time it would take to go from zero to the speed of light (186,000 MPS) at one steady G force. Or where to look for the answer?

### 2 Answers

- statmanLv 61 decade agoFavorite Answer
Since our initial velocity is 0, our equation is

v = gt

where v is velocity, g is acceleration, and t is time. Hence,

t = v/g

Since g is 32 feet/sec/sec, let's convert light speed to feet/sec.

(186,000 miles/sec)(5280 feet/mile) = 982080000 feet/sec

So the time in seconds will be (982080000 feet/sec)/(32 feet/sec/sec)

= 30690000 sec

Divide by 3600 sec/hour to get hours,

= 8525 hours

Divide by 24 hours/day to get days,

= 355.2083 days

So it would take nearly a year!

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- Andy JLv 71 decade ago
One G is 9.81 m/s². Just divide into the speed of light (299,792,458 m/s) to get:

299,792,458 m/s / 9.81 m/s² = 30,559,883 s

That's 353.7 days.

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