Question

49. The displacement of a particle after time t is given by x = (t3 -6+2+31+4) metre. What is
the velocity of the particle when its acceleration is zero
1) - 12 m/s
2) - 9 m/s
3) - 6 m/s
4)-3 m/s​

Answer 1

Question :

The displacement of a particle after time t is given by x = (t³ - 4t²+ 3t) metre. What is

the velocity of the particle,when its acceleration is zero.

Explanation :

Given :

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

To find :

• Velocity of the particle when the acceleration produced by the particle is 0, v = ?

Knowledge required :

• Differentiation of displacement of a particle gives the velocity of that particle.

So the formula for Velocity of a particle is the change of change of position of the particle with change time, i.e, v = d(x)/dt.

• Differentiation of acceleration of yhe particle gives the acceleration of that particle.

So the formula for acceleration of a particle is the change of change of velocity of the particle with change in time, i.e, a = d(v)/dt.

• Differentiation of a constant term is 0,i.e, d(c)/dx = 0, [Where, c is a constant]

• Power rule of differentiation,i.e d(xⁿ)/dx = x⁽ⁿ ⁻ ¹⁾.

Solution :

To find the velocity of the particle :

⠀By using the equation for velocity of a particle and differentiating it w.r.t time, we get :

⠀⠀=> v = d(x)/dt

⠀⠀=> v = d(t³ - 4t² + 3t)/dt

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

⠀⠀=> v = [3 × t⁽³ ⁻ ¹⁾] - [2 × 4t⁽² ⁻ ¹⁾] + [1 × 3t⁽¹ ⁻ ¹⁾]

⠀⠀=> v = [3 × t²] - [2 × 4t¹] + [1 × 3t⁰]

⠀⠀=> v = 3t² - 8t + 3

⠀⠀⠀∴ v = (3t² - 8t + 3) m/s

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

To find the acceleration of the particle :

⠀By using the equation for acceleration of a particle and differentiating it w.r.t time, we get :

⠀⠀=> a = d(v)/dt

⠀⠀=> a = d(3t² - 8t + 3)/dt

⠀⠀=> a = d(3t²)/dt - d(8t)/dt + d(3)/dt

⠀⠀=> a = [2 × 3t⁽² ⁻ ¹⁾] - [1 × 8t⁽¹ ⁻ ¹⁾] + 0

⠀⠀=> a = [2 × 3t¹] - [1 × 8t⁰]

⠀⠀=> a = 6t - 8

⠀⠀⠀∴ a = (6t - 8) m/s²

Hence the acceleration of the particle is (6t - 8) m/s².

To find the instant of time when the acceleration produced by the particle is zero :

⠀By using the acceleration of the particle and substituting the values in it, we get :

⠀⠀=> a = 6t - 8

⠀⠀=> a = 6t - 8

⠀⠀=> 8 = 6t

⠀⠀=> 8/6 = t

⠀⠀=> 1.4 = t

⠀⠀⠀⠀∴ t = 1.4 s

Hence the instant of time when the acceleration produced by the particle is zero is 2 s.

To find the velocity of the particle at t = 2 s.

By using the velocity of the particle and finding it's velocity at, t = 2 s , we get :

⠀⠀=> v = 3t² - 12t + 3

⠀⠀=> v = 3(1.4)² - 12(1.4) + 3

⠀⠀=> v = 5.88 - 16.8 + 3

⠀⠀=> v = -7.92 or - 8(approx.)

⠀⠀⠀∴ v = -8 m/s

Hence,Velocity of the particle when the acceleration produced by the particle is 0, v = -8 m/s

Answer 2

$\Large\bf\pink{GiVeN,}$

• The displacement of a particle after time t is given by $\bf{x\:=\:(t^3\:-\:6t^2\:+\:3t\:+\:4)\:m}$.

$\Large\bf\gray{To\:FiNd,}$

• The velocity of the particle when it's acceleration is zero.

$\Large\bf\blue{CaLcUlAtIoN,}$

$\bf\red{We\:know\:that,}$

$\orange\bigstar\:\:\bf{\color{indigo}Velocity\:=\:\dfrac{Change\:in\:Displacement \:(dx)}{Change\:in\:time\:(dt)}\:}$

$:\implies\:\:\bf{Velocity\:=\:\dfrac{d(t^3\:-\:6t^2\:+\:3t\:+\:4)}{dt}\:}$

$:\implies\:\:\bf{Velocity\:=\:\dfrac{dt^3}{dt}\:-\:\dfrac{d(6t^2)}{dt}\:+\:\dfrac{3t}{dt}\:+\:\dfrac{d (4)}{dt}\:}$

$:\implies\:\:\bf{Velocity\:=\:3t^2\:-\:12t\:+\:3\:+\:0\:}$

$:\implies\:\:\bf\blue{Velocity\:=\:(3t^2\:-\:12t\:+\:3)\:m/s}--(1)$

$\bf\red{Again,}$

$\green\bigstar\:\:\bf\purple{Acceleration \:=\:\dfrac{Change\:in\:Velocity \:(dv)}{Change\:in\:time\:(dt)}\:}$

$:\implies\:\:\bf{Acceleration \:=\:\dfrac{d(3t^2\:-\:12t\:+\:3)}{dt}\:}$

$:\implies\:\:\bf{Acceleration \:=\:\dfrac{d(3t^2)}{dt}\:-\:\dfrac{d(12t)}{dt}\:+\:\dfrac{d (3)}{dt}\:}$

$:\implies\:\:\bf{Acceleration \:=\:6t\:-\:12\:+\:0\:}$

$:\implies\:\:\bf\pink{Acceleration \:=\:(6t\:-\:12)\:m/s^2}$

☆ It is given that acceleration is zero.

$:\implies\:\:\bf{0\:=\:(6t\:-\:12)\:}$

$:\implies\:\:\bf{6t\:=\:12}$

$:\implies\:\:\bf{t\:=\:\dfrac{12}{6}}$

$:\implies\:\:\bf\orange{t\:=\:2\:sec}$

☆ Now put the value of t in equation(1), we get

$\longmapsto\:\:\bf{Velocity\:=\:3\times{2^2}\:-\:12\times{2}\:+\:3}$

$\longmapsto\:\:\bf{Velocity\:=\:3\times{4}\:-\:24\:+\:3}$

$\longmapsto\:\:\bf{Velocity\:=\:12\:-\:21}$

$\longmapsto\:\:\bf\green{Velocity\:=\:-\:9\:m/s}$

$\Large\bf\therefore$ 2) The velocity of the particle is -9m/s.