Question

The velocity 'V' of moving
object varies with
displacement 'x' as v = 3x2 -
2x, then the acceleration at x
= 2m is
​

Answer 1

Given,

Velocity varies with displacement as v = 3x²-2x.

To Find,

The value of acceleration at x = 2m.

Solution,

Since the equation for velocity is given

v = 3x²-2x.

Also,

a = v*dv/dx

So,

a = 3x²-2x d(3x²-2x)/dx

a = 3(2)(2)-2(2){(6(2)-2)]

a = (12-4)(12-2)

a = 8(10)

a = 80 m/s²

Hence, the value of acceleration is 80 m/s².

Answer 2

Given velocity varies with displacement x ,

$v = {3x}^{2} - 2x$ ...(1)

To find Acceleration at x = 2m

solution :

formula to be used ,

$v = \frac{dx}{dt}$

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

=> differentiate equation 1 with respect to t

$\frac{dv}{dt} =( 6x - 2) \frac{dx}{dt}$

$a = (6x - 2)v$

$a = (6x - 2)(3 {x}^{2} - 2x) \: .....(2)$

at x = 2m

on putting value of x in equation 2 we get

$a = (6 \times 2 - 2)(3 \times 4 - 4) = 10 \times 8 = 80 \frac{m }{ {s}^{2} }$

Hence acceleration at x = 2m is 80m/s^2