Question

a body starts with initial velocity along a staight line .after 3 s it is found to be a distance of 15 m from the startung point . at 4s it is at starting point what is the initial velocity of the body assume constant accelearation

Answer 1

Answer:

• The initial velocity (u) of the body will be 20 m/s.

Given:

1. Let the Initial velocity be ' u '
2. Let the constant acceleration be ' a '.

Explanation:

$\rule{300}{1.5}$

Case - 1

From second kinematic equation.

⇒ S = u t + 1 / 2 a t²

Here,

• S Denotes Displacement.
• u Denotes Initial velocity.
• t Denotes time.
• a Denotes Acceleration.

Now,

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ 15 = u × 3 + 1 / 2 × a × (3)²

 ∵ [ S = 15 m, t = 3 sec]

⇒ 15 = 3 u + 1 / 2 × a × 9

Multiplying by 2 on both sides.

⇒ 2 × 15 = 2 × 3 u + 2 ×(1 / 2 × a × 9)

⇒ 30 = 6 u + a × 9

⇒ 30 = 6 u + 9 a

⇒ a = ( 30 - 6 u ) / 9 ____ [ 1 ]

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Case - 2

From second kinematic equation.

⇒ S = u t + 1 / 2 a t²

Here,

• S Denotes Displacement.
• u Denotes Initial velocity.
• t Denotes time.
• a Denotes Acceleration.

Now,

⇒ S = u t + 1 / 2 a t²

Substituting the values,

Since, the particle returns to its Initial position then the Displacement will be Zero.

⇒ 0 = u × 4 + 1 / 2 × a × (4)²  

∵ [ S = 0 m, t = 4 sec]

⇒ 0 = 4 u + 1 / 2 × a × 16

Multiplying by 2 on both sides.

⇒ 0 × 2 = 2 × 4 u + 2 × (1 / 2 × a × 16)

⇒ 0 = 8 u + a × 16

⇒ 0 = 8 u + 16 a

⇒ 16 a = - 8 u

Substituting the value of acceleration from equation [ 1 ]

⇒ 16 × ( 30 - 6 u) / 9 = - 8 u

⇒ ( 30 - 6 u) / 9 = - 8 u / 16

⇒ ( 30 - 6 u) / 9 = - u / 2

Cross - multiplying,

⇒ 2 × ( 30 - 6 u) = 9 × - u

⇒ 60 - 12 u = - 9 u

⇒ 60 = - 9 u + 12 u

⇒ 3 u = 60

⇒ u = 60 / 3

⇒ u = 20

⇒ u = 20 m/s.

∴ The initial velocity (u) of the body will be 20 m/s.

$\rule{300}{1.5}$