Question

please answer this question briefly ​

Answer 1

Given:

A particle moves along a straight line and the curve represents the velocity time graph representing a semicircle with centre on the x-axis.

To find:

Acceleration at x = 17 m?

Calculation:

We know that acceleration can be written as:

$\boxed{ \rm \: a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \times \dfrac{dx}{dt} = v \dfrac{dv}{dx} }$

First, let's write down the general equation for the semicircle.

$\rm \therefore \: {v}^{2} + {(x - 15)}^{2} = {5}^{2}$

Now, differentiation w.r.t x :

$\rm \implies \: \dfrac{d {v}^{2} }{dx} + \dfrac{d{(x - 15)}^{2}}{dx} = \dfrac{ d{5}^{2} }{dx}$

$\rm \implies \: 2v \dfrac{dv}{dx} + 2(x - 15) = 0$

$\rm \implies \: 2v \dfrac{dv}{dx} = - 2(x - 15)$

$\rm \implies \: v \dfrac{dv}{dx} = - x + 15$

$\rm \implies \: a = - 17 + 15$

$\rm \implies \: a = - 2 \: m/ {s}^{2}$

So , acceleration is -2 m/s² at x = 17 m.