Question

The velocity-position graph of a particle moving in a straight line along x-axis is given below. Acceleration of particle at x = 2 m is v (m/s) v= 2x x (m) 0 (1) 8 m/s2 (3) 64 m/s2 (2) 16 m/s2 (4) 32 m/s2​

Answer 1

Given:-

• V= 2x²

To find:-

• Acceleration at x=2m

Solution:-

For finding acceleration at 2m we need to find acceleration as function of distance

Using differentiation

$\large\rightarrow \: \: \frac{dv}{dt} = a$

but Velocity is given as function of distance so we can not differentiate it hence we need to change the appearance.

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

$\large \: \rightarrow \: \: \frac{d(2 {x}^{2} )}{dx} \times \frac{dx}{dt} = a$

Now we know that $\frac{dx}{dt}=v$

Hence

$\large \: \rightarrow \: \: \frac{d(2 {x}^{2}) }{dx} \times v = a$

$\large \: \rightarrow \: \: \frac{d(2 {x}^{2}) }{dx} \times 2 {x}^{2} = a$

$\large \: \rightarrow \: \: 4x \times 2 {x}^{2} = a$

$\large \: \rightarrow \: \: 8 {x}^{3} = a$

Now we have got acceleration as function of distance

Hence at 2m

acceleration=8(2)³=64m

$\huge\mathfrak\red{\boxed {answer - 64m}}$