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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

  • 1 decade ago
    Favorite 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 J
    Lv 7
    1 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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