Question

A particle moving with a uniform acceleration travels 24 metre and 64 metre in first two successive intervals of 4 second each. Its initial velocity is

Answer 1

Let initial velocity (u)

A= 4u + 8a=24

4u + 16a + 8a=64

a=40/16

▶u=1 m/s...Ans✔

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

D traveled in 4 sec.

= 24 m

Another 4 sec. = 64 m

Time taken (t) = 4 sec.

Assume,

Initial velocity of particle = u m/s.

Now,

Using 1st Equation of motion,

v = u + at

⇒ v = u + a × 4

⇒ v = u + 4a

Also,

Using 2nd Equation of motion,

⇒ s = ut + 1/2 at²

⇒ 24 = 4u + 1/2 × a × 4 × 4

⇒ 24 = 4u + 8a

⇒ u + 2a = 6 ......... (1)

Also, For another 4 sec,

⇒ s = ut + 1/2 at²

Using (1) we have,

⇒ 6u = (u + 4a) × 4 + 1/2 × a × 16

⇒ 16 = u + 4a + 2a

⇒ u + 6a = 16 ........... (2)

Now,

From (1) & (2) we have,

⇒ 4a = 10

⇒ a = 10/4

⇒ a = 2.5

Now,

Substitute value of a in (1),

⇒ u + 2a = 6

⇒ u + 2 × 2.5 = 6

⇒ u + 5 = 6

⇒ u = 6 - 5

⇒ u = 1 m/s

Therefore,

Initial velocity of particle = 1 m/s