# uniform acceleration

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Thread starter 2 years ago
#1
One mark answer question I can't get my head around:
Car accelerates uniformly from rest along a straight road (negligible frictional force) the car travels 15m in the 2nd second- how far does it travel in the fifth second?

I did;
in the first second it would have travelled 7.5m
so total distance travelled by the second second= 22.5m
2s/t^2=a 2x22.2/22= 11.25ms^-2
Then I calculated the total distance travelled at the 5th and 4th second;
S=1/2 at ^2 5th; 0.5 x 11.25 x 5^2= 140.6m.
4th ; 0.5 x 11.25 x 4^2= 90m
5th second total distance – 4th total second distance= 51m

The actual answer is 45m and I hate my life.
Anyone know where I went wrong and if there’s a simper way?
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2 years ago
#2
(Original post by Nikki.Dobson)
in the first second it would have travelled 7.5m

How did you work that out?

I'd say, in the first second it travels s = 0.5a*1²

and by the end of the second it has travelled 0.5a*2² so the difference between these is 15m

In other words, 0.5 a * (4² - 1²) = 15 Edited: Should of course be (2² - 1²)

That will allow you to find a then you should be able to do the rest.
1
Thread starter 2 years ago
#3
I’m guessing you meant ‘2’ when you said ‘4’? but thank you sooooo much that works! That took me like an entire 2 hours to understand. :’)
0
2 years ago
#4
(Original post by Nikki.Dobson)
I’m guessing you meant ‘2’ when you said ‘4’? but thank you sooooo much that works! That took me like an entire 2 hours to understand. :’)
Oh, sorry!! Yes, 2² = 4... It was late at night. Have edited it in case it confuses anyone else

Glad you got it though.
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