Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Answers
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 will be 9/2a.
An object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Let us assume that the initial velocity of the object is 'u', final velocity of the object is 'v', acceleration is 'a' and is moving in time 't'.
For 4th second:
Using the Second Equation Of Motion,
s = ut + 1/2 at²
s = u(4) + 1/2 a(4)²
s = 4u + 1/2 × a × 16
s = 4u + 8a ................(1st equation)
Similarly for 5th second:
Using the Second Equation Of Motion,
s' = ut + 1/2 at²
s' = u(5) + 1/2 a(5)²
s' = 5u + 1/2 × a × 25
s' = 5u + 25a/2 .............(2nd equation)
We have to find the distance travelled by the object.
So,
Distance travelled by object = Distance travelled by object in 5 seconds - Distance travelled by object in 4 seconds
Substitute the known values from above calculations,
Distance travelled by object = 5u + 25a/2 - (4u + 8a)
= 5u + 25a/2 - 4u - 8a
= 5u - 4u + 25a/2 - 8a
= u(5 - 4) +a(25/2 - 8)
= u(1) +a[(25 - 16)/2]
= u +a(9/2)
= u + 9a/2
Therefore, the distance travelled by an object is (u + 9a/2)m moving with a uniform acceleration in the interval between 4th and 5th seconds.