Question

how I can intereste it●

Answer 1

Hey There!!!


Here, we are given two pieces of information:

$a=3x^2$
And, v = 2 m/s when x=0


We are given acceleration as a function of position. Now, we also need to derive velocity as a function of position. Here's how we are going to do it.


$a = \frac{dv}{dt} \\ \\ \\ \text{Multiply and divide by dx} \\ \\ \\ \implies a = \frac{dv}{dx} \times \frac{dx}{dt} \\ \\ \\ \implies a = \frac{dv}{dx} \times v \\ \\ \\ \implies v \frac{dv}{dx} = a \\ \\ \\ \implies v \frac{dv}{dx} = 3x^2 \\ \\ \\ \implies v \, dv = 3x^2 \, dx \\ \\ \\ Integrating \\ \\ \\ \implies \int v \, dv = \int 3x^2 \, dx \\ \\ \\ \implies \frac{v^2}{2} = 3 \times \frac{x^3}{3} + c \\ \\ \\ \implies v^2 = 2x^3 + 2c$

Here, 2c is the constant of integration. We may as well write 2c as simply c.

So,

$v^2 = 2x^3+c$


Now, we will use the information that v=2 when x=0:

$2^2 = 0 + c \\ \\ \\ \implies c = 4$


Thus, now we finally have velocity as a function of position:


$v^2 = 2x^3 + c \\ \\ \\ \implies v^2 = 2x^3 + 4 \\ \\ \\ \implies v = \sqrt{2x^3+4}$

Now, we are asked to find the velocity when  x = 4 m.

$v = \sqrt{2(4)^3+4} \\ \\ \\ \implies v = \sqrt{2(64)+4} \\ \\ \\ \implies v = \sqrt{132} \\ \\ \\ \implies \boxed{v = 2\sqrt{33} \, \, m / s}$


Hope it helps
Purva
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