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# uniform acceleration

Wed 4 May, 2011 09:01 am
i got a question in my physics test that a body is moving from rest with 4m/s² uniform acceleration. Find distance covered by it in 10 secs.

does it mean that its velocity is increasing 4m/s in every sec?
If so then, in 1st sec it moves with 4m/s,2nd-8m/s,3rd-12m/s and so on till ten seconds...
So, distance covered = 4+8+12+16+20+24+28+32+36+40 =220m

But by equaton of motion... we have
S=ut + 1/2 at²
S= (0)(10)+1/2*(4)(10)²
S=0+200
so,distance travelled = 200m

BY BOTH THESE DIFFERENT METHODS,I'M GETTING DIFFERENT ANSWERS.....CAN YOU PLEASE TELL ME WHERE AM I WRONG??
THANKING YOU
JATIN
CLASS-9th
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Wed 4 May, 2011 10:35 am
Its velocity increases 4m/s every sec.

So during the first second, it starts with a velocity of zero, ends with a velocity of 4m/s. Average velocity during that second is 2m/s.

At the end of one second: distance traveled is 2 meters, ending velocity is 4m/2

check by equation:

d(t)=1/2at^2 +vt
d(1) =1/2(4m/s^2)*(1s)^2 +(0)(1s)
d(1)=2m

During the second second of motion, it starts with a velocity of 4m/2, ends with a velocity of 8m/s, for an average velocity of 6m/s. Distance during the second second is 6m, for a total distance of 8m.

Check by equation:

d(t)=1/2at^2 + vt

d(2) = 1/2 (4m/s^2) * (2s)^2+0(2s)
d=1/2(4)(4)m = 8m

At the end of 10 seconds:

d(10) = 1/2(4m/s^2)*(10s)^2 + 0(10s)
d(10) = 1/2(4m/s^2)*100s^2
d(100)=200m
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