Question

. A particle cover 16m during 4th second of its motion and 24m during 6th second of its motion.

Determine the value of initial velocity as well as acceleration.



please give the correct answer otherwise your answer will be reported❤❤❤❤​

Answer 1

Answer:

Answer :

Acceleration of the particle, a = 4 m/s²

Initial velocity of the particle, u = 2 m/s

Explanation :

Given :

Distance covered by the particle during 4th second of it's motion, s₄ = 16 m

Distance covered by the particle during 6th second of it's motion, s₆ = 16 m

To find :

Acceleration of the particle, a = ?

Initial velocity of the particle, u = ?

Knowledge required :

Formula for distance covered in nth second of a body :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Sn = u + ½a(2n - 1)⠀

[Where : Sn = Distance traveled by the body in it's nth second, a = Acceleration of the body, n = nth second, u = Initial velocity of the body]

Solution :

Equation for the motion of the particle in it's 4th s of it's motion :

By using the formula nth second of a particle, we get :

⠀⠀⠀=> Sn = u + ½a(2n - 1)

⠀⠀⠀=> 16 = u + ½a(2(4) - 1)

⠀⠀⠀=> 16 = u + ½a(8 - 1)

⠀⠀⠀=> 16 = u + ½a × 7

⠀⠀⠀=> 32 = 2u + 7a⠀⠀⠀⠀⠀⠀⠀⠀...(i)

Hence the equation for the motion of the particle in it's 4th s of it's motion is (32 = 2u + 7a).

Equation for the motion of the particle in it's 4

6th s of it's motion :

By using the formula nth second of a particle, we get :

⠀⠀⠀=> Sn = u + ½a(2n - 1)

⠀⠀⠀=> 24 = u + ½a(2(6) - 1)

⠀⠀⠀=> 24 = u + ½a(12 - 1)

⠀⠀⠀=> 24 = u + ½a × 11

⠀⠀⠀=> 48 = 2u + 11a⠀⠀⠀⠀⠀⠀⠀⠀...(ii)

Hence the equation for the motion of the particle in it's 6th s of it's motion is (48 = 2u +11a).

Now by subrre5 Eq.(i) and Eq.(ii), we get :

⠀⠀⠀=> 32 - 48 = (2u + 7a) - (2u + 11a)

⠀⠀⠀=> -16 = 2u + 7a - 2u - 11a

⠀⠀⠀=> -16 = -4a

⠀⠀⠀=> 4 = a

⠀⠀⠀⠀∴ a = 4 m/s²

Now by substituting the value of a in the Eq.(i),we get :

⠀⠀⠀=> 32 = 2u + 7a

⠀⠀⠀=> 32 = 2u + 7(4)

⠀⠀⠀=> 32 = 2u + 28

⠀⠀⠀=> 32 - 28 = 2u

⠀⠀⠀=> 2 = u

⠀⠀⠀⠀∴ a = 2 m/s

Therefore,

Acceleration of the particle is 4 m/s².

Initial velocity of the particle is 2 m/s.

Answer 2

Answer:

At 4ᵗʰ second distance covered ﹦ 16 m

=﹥ speed at that moment = 16m/s

At 6ᵗʰ Second distance covered = 24m

=﹥ speed at that moment = 24 m / s

By Equation of motion,

v=u+at

=>24=16 + a x 2

=>8=2a

=>a=4m/s²

Initial Velocity = 2m/s

Acceleration= 4m/s²