Question

A body moving with a uniform acceleration travels distances of 24 m and 64 m during the
first two equal consecutive intervals of time, each of duration 4 s. Determine the initial
velocity and acceleration of the moving body.
Ans: 1m/s ,2.5 m/s2​

Answer 1

Answer:

answer is 1m/s ,2.5 m/s2

Answer 2

Given :-

Distance covered by the body in the first interval of time = 24 m

Distance covered by the body in the second interval of time = 64 m

Time taken in the first two equal consecutive intervals of time = 4 sec

To Find :-

The initial  velocity of the body.

The acceleration of the body.

Solution :-

We know that,

• t = Time
• s = Distance
• u = Initial velocity
• a = Acceleration
• v = Final velocity

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Time (t) = 4 sec

Substituting their values,

⇒ v = u + 4a

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} at^2}}$

Given that,

Distance (s) = 24 m

Time (t) = 4 sec

Substituting their values,

⇒ 24 = 4u + 1/2 × a × 4²

⇒ 24 = 4u + 1/2 × a × 16

⇒ 24 = 4u + 16a / 2

⇒ 24 = 4u + 8a

⇒ u + 2a = 6   ....(1)

For the next 4 sec,

Given that,

Initial velocity (u) = u + 4a

Time (t) = 4 sec

Distance (s) = 64 sec

Substituting their values,

⇒ 64 = (u + 4a) × 4 + 1/2 × a × 16

⇒ 64 = (u + 4a) × 4 + 16a

⇒ 16 = u + 4a + 2a

⇒ 16 = u + 6a    ....(2)

From equation (1) and (2),

(u + 6a = 16) - (u + 2a = 6)

= 4a = 10

a = 10/4

a = 2.5 m/s

Therefore, the acceleration of the moving body is 2.5 m/s.

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Final velocity (v) = 6 m/s

Acceleration (a) = 2.5 m/s

Substituting their values,

⇒ 6 = u + 2.5 × 2

⇒ 6 = u + 5

⇒ u = 6 - 5

⇒ u = 1 m/s

Therefore, the initial velocity of the moving body is 1 m/s.