A body covers 12 m in 2 nd s and 20 m in4 th s.how much distance is covered in 4 s after the 5 th second
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Given that the motion is in 1 Dimension and the body doesn't change it's direction (i.e. velocity is in the direction of acceleration), we have:
s=u+a(2t−1)/2s=u+a(2t−1)/2
12 = u + a(2*2 - 1)/2
=> 12 = u + 3a/2
=> 2u + 3a = 24
and
20 = u + a(2*4 - 1)/2
20 = u + 7a/2
=> 2u + 7a = 40
From 1 and 2:
a = 4
u = 6
Now, distance covered in 4s after 5th second = Distance covered in 9 seconds - distance covered in 5 seconds
= 6*9 + 1/2 * 4 * 9 * 9 - (6 * 5 + 1/2 * 4 * 5 * 5)
= 54 + 162 - (30 + 50)
= 136 m
s=u+a(2t−1)/2s=u+a(2t−1)/2
12 = u + a(2*2 - 1)/2
=> 12 = u + 3a/2
=> 2u + 3a = 24
and
20 = u + a(2*4 - 1)/2
20 = u + 7a/2
=> 2u + 7a = 40
From 1 and 2:
a = 4
u = 6
Now, distance covered in 4s after 5th second = Distance covered in 9 seconds - distance covered in 5 seconds
= 6*9 + 1/2 * 4 * 9 * 9 - (6 * 5 + 1/2 * 4 * 5 * 5)
= 54 + 162 - (30 + 50)
= 136 m
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