obtain a relation for distance travelled by an object moving with a uniform accelaration in the interval between 4th and 5th seconds
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638
Uniform acceleration in the interval between 4th and 5th seconds =
= (5u + 1/2 * a * 5^2) - (4u + 1/2 * a * 4^2)
= 5u + 1/2 *a * 25 - 4u + 1/2 * a * 16
= u + 1/2 * a * (25-16)
= u + 1/2 * a * 9
= u + 9/2 * a
If it starts from rest then the displacement will be 9/2a.
Hope this helps!
= (5u + 1/2 * a * 5^2) - (4u + 1/2 * a * 4^2)
= 5u + 1/2 *a * 25 - 4u + 1/2 * a * 16
= u + 1/2 * a * (25-16)
= u + 1/2 * a * 9
= u + 9/2 * a
If it starts from rest then the displacement will be 9/2a.
Hope this helps!
mdfarooq12335:
S5-S4
Answered by
28
Answer:
Explanation:Uniform acceleration in the interval between 4th and 5th seconds =
= (5u + 1/2 * a * 5^2) - (4u + 1/2 * a * 4^2)
= 5u + 1/2 *a * 25 - 4u + 1/2 * a * 16
= u + 1/2 * a * (25-16)
= u + 1/2 * a * 9
= u + 9/2 * a
If it starts from rest then the displacement is u+9/2a.
Hope this helps you
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