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?
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?
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.
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. :’)
Glad you got it though.