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Question

a body moving with uniform accelerationdescribes 20m in 2nd second and 30m in 4th second of its motion.how to calculate the distance moved by the body in 6th second.

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Solution

s = u + (0.5)a(2n-1)
where s is the distance travelled in nth second.
u is the initial velocity.
n is the nth second


Just put the values!
20 = u + 0.5 * a* (2*2 - 1)
20 = u + 0.5 * a* 3
20 =
u + 1.5 a..........(1)

for the second case:
30 = u+ 0.5(a) (2*4-1)
30 =
u+ 0.5(a) (7)
30 = u+ 3.5a.........(2)

use equation 1 and 2 to get "u" and "a"
a = 2m/s2
u = 17m/s

now just put the values in the same formula for the nth second and get the answer that is
a = 5m/sec2
v=12.5m/s




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