Question

For motion of an obiect along the x-axis the
velocity v and displacement x are related as
V= 3x2- 2x. The acceleration of object at x = 2 mis
(1) 48 m/s2
(2) 80 m/s2
(3) 18 m/s2
4 10 m/s2​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{\bold{V = 3x^2 - 2x}}$

a = Acceleration.

v = Velocity.

t = Time taken

$\huge{\bold{\underline{Explanation:-}}}$

As You Know,

$\large{\bold{a = \frac{dv}{dt}}}$

Now, rewriting.

$\large{a = \frac{dv}{dt} \times \frac{dx}{dx}}$

it becomes,

$\large{a = v. \frac{dv}{dx} ------(1)}$

Substituting the values.

$\large{ a = (3x^2 - 2x) \times \frac{d (3x^2 - 2x)}{dt}}$

$\large{ a = (3x^2 - 2x) \times (6x - 2)}$

As x = 2 m is given then,

$\large{ a = (3 \times {2}^2 - 2 \times 2 ) \times {(6 \times 2 - 2)}}$

$\large{ a = (3 \times 4 - 2 \times 2 ) \times {(12 - 2)}}$

$\large{ a = (12 - 4) \times 10}$

$\large{ a = 8 \times 10}$

$\huge{\boxed{\boxed{a = 80 \: m/s^2}}}$

So,the acceleration of the object is 80 m/s².

Answer 2

$\huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Velocity of the particle is defined as:

$\mathsf{v = 3{x}^{2} - 2x }$

To find

Acceleration of the particle at x = 2m

For this,

We need to differentiate "v" w.r.t to "x",so that we get acceleration

We Know that,

$\huge{\boxed{ \sf{a = v. \frac{dv}{dx} }}}$

Putting the values,we get:

$\sf{a = (3x{}^{2} - 2x) \frac{d(3x {}^{2} - 2x) }{dx} } \\ \\ \rightarrow \: \sf{a = (3 {x}^{2} - 2x)(6x - 2) } \\ \\ \rightarrow \sf{a = 18x{}^{3} - 12x {}^{2} - 6x {}^{2} + 4x} \\ \\ \huge{ \rightarrow \: \sf{a = 18x {}^{3} - 18x {}^{2} + 4x }}$

Putting x = 2,we obtain:

$\sf{a = 18(2) {}^{3} - 18(2) {}^{2} + 4(2) } \\ \\ \implies \: \huge{\sf{a = 80 \: ms {}^{ - 2} }}$

Thus,the acceleration of the particle at x = 2 is 80m/s²