Question

A particle moves along a straight line such that its displacement at any time t is given by s= (t^3-6t^2+3t+4) meters. What is the velocity when acceleration is zero?​

Answer 1

Answer :

• Velocity of the particle when accelaration is (-9) m/s.

Explanation :

Given :

• Displacement of the particle, s = (t³ - 6t² + 3t + 4) m.
• Acceleration of the particle, a = 0 m/s².

To find :

• Velocity of the particle when accelaration is 0, v = ?

Knowledge required :

• Differentiation of the displacement of the body w.r.t time gives the velocity of the body.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀v = d(s)/dt

• Differentiation of the velocity of the body w.r.t time gives the accelaration of the body.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀a = d(v)/dt

• Exponent rule of differentiation , x^n = nx^(n - 1)

• Differentiation of a constant term is zero, d(c)/dx = 0

Solution :

To find the velocity of the particle :

We know that if we differentiate the displacement of the particle , we will get the velocity of that particle.

By differentiating the displacement of the particle w.r.t t, we get :

⠀⠀=> v = d(s)/dt

⠀⠀=> d(s)/dt = d(t³ - 6t² + 3t + 4)/dt

⠀⠀=> d(s)/dt = d(t³)/dt - d(6t²)/dt + d(3t)/dt + d(4)/dt

⠀⠀=> d(s)/dt = (3)t⁽³ ⁻ ¹⁾ - (2)6t⁽² ⁻ ¹⁾ + (1)3t⁽¹ ⁻ ¹⁾ + 0

⠀⠀=> d(s)/dt = 3t² - 12t + 3

⠀⠀⠀⠀⠀⠀⠀∴ v = (3t² - 12t + 3) m/s

Hence the velocity of the particle is (3t² - 12t + 3) m/s.

To find the acceleration of the particle :

We know that if we differentiate the velocity of the particle , we will get the acceleration of that particle.

By differentiating the velocity of the particle w.r.t t, we get :

⠀⠀=> a = d(v)/dt

⠀⠀=> d(v)/dt = d(3t² - 12t + 3)/dt

⠀⠀=> d(v)/dt = d(3t²)/dt - d(12t)/dt + d(3)/dt

⠀⠀=> d(v)/dt = (2)3t⁽² ⁻ ¹⁾ - 12t⁽¹ ⁻ ¹⁾ + 0

⠀⠀=> d(v)/dt = (6t - 12)

⠀⠀⠀⠀⠀⠀⠀∴ a = (6t - 12) m/s²

Hence the accelaration of the particle is (6t - 12) m/s².

According to the question, the accelaration of the particle is 0.

By substituting the value of a in the accelaration of the particle and solving the equation, we get :

⠀⠀=> a = (6t - 12)

⠀⠀=> 0 = 6t - 12

⠀⠀=> 12 = 6t

⠀⠀=> 12/6 = t

⠀⠀=> 2 = t

⠀⠀⠀⠀⠀⠀⠀∴ t = 2 s.

Hence the in instant of time when acceleration is 0 is 2 s.

According to the question, we have to find the velocity of the particle when the accelaration is zero, hence the instant of time will be same.

Now let's find out the velocity of the particle at t = 2 s.

By substituting the value of t in the velocity of the particle, we get :

⠀⠀=> v = 3t² - 12t + 3

⠀⠀=> v₍ₜ ₌ ₂ ₛ₎ = 3(2)² - 12(2) + 3

⠀⠀=> v₍ₜ ₌ ₂ ₛ₎ = 3(4) - 12(2) + 3

⠀⠀=> v₍ₜ ₌ ₂ ₛ₎ = 12 - 24 + 3

⠀⠀=> v₍ₜ ₌ ₂ ₛ₎ = -9

⠀⠀⠀⠀⠀⠀⠀∴ v₍ₜ ₌ ₂ ₛ₎ = -9 m/s

Therefore,

• Velocity of the particle when accelaration is 0, v = (-9) m/s.