if a particle starts moving with acceleration 2m/s2 . what is th distance travelled by it in 5th half second ?

Here acc. is 2m/s2 & 5th half means 2.5 so

1/2a(2.5)2 - 1/2a(2)2

=1/2 *2(6.25-4)

=2.25

  • 99
41.25m
  • -21
where did u see this question??????
  • -18
in aakash package
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Which formula u have used here
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Please tell why subtract
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Refer to the picture

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Why have u taken initial time 2 sec??
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a=2 ms-2
n=5 x ½=5/2
so, x5/2=½ x a x (2n-1)
=2 x 5/2-1
=4m
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I think, it would be more clear...
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Ans to the question

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Please find this answer

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Please find this answer

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find the displacement in time 2second to 7second
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ihbukj
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Use s=1/2at^2 because initial speed is zero And time taken is 2 and 2.5 and solve it like this s=1/2a(2.5*2.5-2*2) s=1/2a(6.25) s=2.25
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2.25m
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Distance travelled in 5th half second = Distance travelled in 2.5 seconds - Distance travelled in 2 seconds
= 0.5a (t₁² - t₂²)
= 0.5 * 2 * (2.5² - 2²)
= 2.25m
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