A body starts from rest and moves with a uniform acceleration of 6 ms \(^{-2}\). What distance does it cover in the third seconds
15m
18m
27m
30m
Explanation
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Discussions (27)
Lol, it's not wrong, it's actually correct
When looking for the distance between seconds
For example the third seconds, we'll use S(3rd second) =ut+ 1/2a(t2-t1)²
That's
S=0+1/2(6)(3²-2²)
To find the third second we'll use the second before it, 2nd second
so
S=3(9-4)
S=3*5
S= 15m 
For those who are still confused.
If in 2 days, a man travels 10km and in three days he travels 20km, the actual distance that he traveled within the third day is 20km - 10km i.e 10km.
This means that within the third day, the man travels an additional 10km.
Also, in the second second i.e 2s
S=ut+½at²
S = ½(6)(2²)
S = 3×4
S = 12m
In the third second i.e 3s
S = 27m (calculated in the same way)
Therefore, the actual distance moved within the third second is 27 - 12
= 15m
S=0+1/2(6)(3²-2²)
To find the third second we'll use the second before it, 2nd second
so
S=3(9-4)
S=3*5
S= 15m
S= ut + 1/2 at²
ut= 0
Distance covered in 3rd or third second is
S= 1/2 x 6 (3²-2²)
S=1/2 x 6 (9-4)
S= 1/2 x 6 x (5)
S=3 x 5
S= 15m
yoi can also use
Sn=a/2(2t-1)
n=2
a=6
t=3
Sn=6/2(2(3)-1)
Sn=3(6-1)
Sn=3(5)
Sn=15m 
You're welcome.
The selected answer is wrong:
a=6ms^(-2)
u=0ms^(-1)
t=3s
S=ut+(1/2)at^2
S=0+(1/2)*6*3^2
S=27m answer
REF: check 3rd equation of motion.
