Question

Q.15) Obtain a relation for the
distance travelled by an object
moving with a uniform acceleration in
the interval between 4th and 5th
seconds. *
O A) u+9a/2
O B)4u+8a
O c)15u+25a/2
OD)none of these​

Answer 1

Answer:

A) u + 9a/2

Explanation:

Distance travelled in 4 sec = S4

S4 = u×4+1/2 × a × 4 × 4

S4 = 4u + 8 a

Distance travelled in 5 sec = S5

S5 = u × 5 + 1/2 × a × 5 × 5

S5 = 5u + 25a / 2

Distance travelled between 4th and 5th sec

5u + 25a /2 - ( 4u + 8a )

= 5u +25a / 2 - 4u - 8a

= u + 9a /2