Question

If position of particle is given by x= t^3 , find acceleration of the particle

Answer 1

Given :

• Position of the particle, x = t³

To find :

• Acceleration of the particle, a = ?

Knowledge required :

• By Differentiating the position of the particle we will get the velocity of the Particle and if we Differentiate the velocity of the Particle, we will get the Acceleration of the particle.

• First principle of derivative :

$\boxed{\sf{f(x)' = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}}}$

• (a + b)³ = a³ + b³ + 3a²b + 3ab²
• (a + b)² = a² + 2ab + b²

Solution :

First let us find the velocity of the particle.

By using the first principle of derivatives and substituting the values in it, we get :

$:\implies \sf{f(x)' = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}}$

Here, we Differentiate with respect to t.

So the equation will be :

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{f(t + h) - f(t)}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{(t + h)^{3} - t^{3}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{t^{3} + 3t^{2}h + 3h^{2}t + h^{3} - t^{3}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3t^{2}h + 3h^{2}t + h^{3}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} 3t^{2} + 3ht + h^{2}}$

$:\implies \sf{f(t)' = 3t^{2} + 3(0)t + (0)^{2}}$

$:\implies \sf{f(t)' = 3t^{2}}$

$\boxed{\therefore \sf{f(t)' = 3t^{2}}}$

Hence the velocity of the Particle is , dv = 3t².

Now let us Differentiate the velocity of the particle :

By using the first principle of derivatives and substituting the values in it, we get :

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{f(t + h) - f(t)}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3(t + h)^{2} - 3t^{2}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3(t^{2} + 2 \times t \times h + h^{2}) - 3t^{2}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3t^{2} + 6th + 3h^{2} - 3t^{2}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{6th + 3h^{2}}{h}}$

$:\implies \sf{f(t)' = \lim_{h \to 0} 6t + 3h}$

$:\implies \sf{f(t)' = 6t + 3(0)}$

$:\implies \sf{f(t)' = 6t}$

$\boxed{\therefore \sf{f(t)' = 6t}}$

Hence the accelaration of the body is 6t m/s².

Answer 2

Explanation:

Given :

Position of the particle, x = t³

To find :

Acceleration of the particle, a = ?

Knowledge required :

By Differentiating the position of the particle we will get the velocity of the Particle and if we Differentiate the velocity of the Particle, we will get the Acceleration of the particle.

First principle of derivative :

\boxed{\sf{f(x)' = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}}}

f(x)

′

=

h→0

lim

h

f(x+h)−f(x)

(a + b)³ = a³ + b³ + 3a²b + 3ab²

(a + b)² = a² + 2ab + b²

Solution :

First let us find the velocity of the particle.

By using the first principle of derivatives and substituting the values in it, we get :

\begin{gathered}:\implies \sf{f(x)' = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}} \\ \\ \\ \end{gathered}

:⟹f(x)

′

=

h→0

lim

h

f(x+h)−f(x)

Here, we Differentiate with respect to t.

So the equation will be :

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{f(t + h) - f(t)}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

f(t+h)−f(t)

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{(t + h)^{3} - t^{3}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

(t+h)

3

−t

3

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{t^{3} + 3t^{2}h + 3h^{2}t + h^{3} - t^{3}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

t

3

+3t

2

h+3h

2

t+h

3

−t

3

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3t^{2}h + 3h^{2}t + h^{3}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

3t

2

h+3h

2

t+h

3

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} 3t^{2} + 3ht + h^{2}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

3t

2

+3ht+h

2

\begin{gathered}:\implies \sf{f(t)' = 3t^{2} + 3(0)t + (0)^{2}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=3t

2

+3(0)t+(0)

2

\begin{gathered}:\implies \sf{f(t)' = 3t^{2}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=3t

2

\begin{gathered}\boxed{\therefore \sf{f(t)' = 3t^{2}}} \\ \\ \\ \end{gathered}

∴f(t)

′

=3t

2

Hence the velocity of the Particle is , dv = 3t².

Now let us Differentiate the velocity of the particle :

By using the first principle of derivatives and substituting the values in it, we get :

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{f(t + h) - f(t)}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

f(t+h)−f(t)

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3(t + h)^{2} - 3t^{2}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

3(t+h)

2

−3t

2

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3(t^{2} + 2 \times t \times h + h^{2}) - 3t^{2}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

3(t

2

+2×t×h+h

2

)−3t

2

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{3t^{2} + 6th + 3h^{2} - 3t^{2}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

3t

2

+6th+3h

2

−3t

2

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} \dfrac{6th + 3h^{2}}{h}} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

h

6th+3h

2

\begin{gathered}:\implies \sf{f(t)' = \lim_{h \to 0} 6t + 3h} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=

h→0

lim

6t+3h

\begin{gathered}:\implies \sf{f(t)' = 6t + 3(0)} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=6t+3(0)

\begin{gathered}:\implies \sf{f(t)' = 6t} \\ \\ \\ \end{gathered}

:⟹f(t)

′

=6t

\begin{gathered}\boxed{\therefore \sf{f(t)' = 6t}} \\ \\ \\ \end{gathered}

∴f(t)

′

=6t

Hence the accelaration of the body is 6t m/s².