Question

obtain a relation for distance travelled by an object moving with a uniform accelaration in the interval between 4th and 5th seconds

Answer 1

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!

Answer 2

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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