Question

A particle is at x=5m at t=2s and has a velocity v=10m/s. It's acceleration is constant at -4m/s². Find the position and velocity at t=1 s and t=3s.​

Answer 1

Answer:

Explanation:

accln a = -4 m/s²

At t= 2s

x₂ = 5 m     and v₂ = 10 m/s

Let x₁ be the position and v₁ be the velocity at t = 1 s

For motion between 1 s to 2 s :

t₂ -t₁ =1 s

v = v + at

v₂ = v₁ + a (t₂-t₁)

10  = v₁ - 4 x 1

v₁ = 14 m/s

s = ut + 1/2 at²

(x₂ - x₁) = v₁(t₂-t₁) + 1/2 a (t₂-t₁)²

5 - x₁ = (14 x 1) -(1/2 x 4 x 1²)

5 - x₁ = 14 -2

∴ x₁ = 5 - 12 = -7 m

∴ At t = 1 s,

position of the particle is ∴ x = -7 m  [at distance of 7 m on negative side of x axis] and velocity is 14 m/s

For motion between 2 s to 3 s :

t₃ -t₂ =1 s

v = v + at

v₃ = v₂ + a (t₃-t₂)

10  = v₂ - 4 x 1

v₂ = 14 m/s

s = ut + 1/2 at²

(x₃ - x₂) = v₂(t₃-t₂) + 1/2 a (t₃-t₂)²

x₃ - 5 = (14 x 1) -(1/2 x 4 x 1²)

x₃ - 5 = 14 -2

∴ x₁ = 12 + 5 = 17 m

∴ At t = 3s,

position of the particle is ∴ x = 17 m  

Thus the particle is exucuting SHM about mean posiyion at time t = 2s