Monday

July 28, 2014

July 28, 2014

Posted by **Sarah** on Sunday, September 10, 2006 at 4:29pm.

a)If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0x 10^8 m/s?

b) How far will it travel in so doing?

I will be happy to critique your work.

I couldn't figure out which equation I should use.

You should see that

s(t)=(1/2)a*t^2 and v(t)=a*t

a is 9.8m/s^2

Find the time t needed to = .1c with the second equation and use that for t in the first one. Please post work

v(t) = at

.1c= 9.8m/s^2*t

t=.0102s

when I used this t for the first equation I got .000509 butt my book siad the answer was 3.1 x 10^&6 s = 1.2 months

No, c=3.0x 10^8 m/s so .1c=3.0x 10^7 m/s. Thus

3.0x 10^7 m/s=9.8m/s^2*t and t=3.0x 10^7 m/s/9.8m/s^2

You should not get t=.0102s

I got t=3061224.49 and when I pluged t into the second equation I got 4.5918 as the answer

The time looks correct. Keep in mind we only had 2 significant figures to begin with. The distance looks ok too, but I think you're missing an exponent of 10.

It should look like 4.6x10^?

What's the ?(something) here?

would it be 4.6x10^6 ?

Now quit guessing. :)

We know s=(1/2)at^2, v=a*t, a=9.8, c=3x10^8

.1c=3x10^7=9.8*t tells us t is approx. 3x10^6

Thus s=approx. 5x(3x10^6)^2=5*9x10^12=45x10^12 or 4.5x10^13

So it can't possibly be 10^6

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