Question

4.
Velocity of an object varies with position x as v=x^2?
what is the acceleration of the object where slope
of v-x curve is equal to one
1)2m/s
2)1/4 m/s
3)1 m/s
4)1/2 m/s​

Answer 1

Answer:

Option (2) 1/4 m/s

Explanation:

Velocity varies as

$v=x^2$

Slope of the above curve = $\frac{dv}{dx}$

$\frac{dv}{dx}=2x$

$\frac{dv}{dx}=1$ at $x=\frac{1}{2}$

At $x=1/2$

$v=\frac{1}{4}$

We know that acceleration is given by

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

$a=v\times 2x$

$\implies a\Bigr|_{x=1/2}=\frac{1}{4}\times 2\times \frac{1}{2}$

$\implies a\Bigr|_{x=1/2}=\frac{1}{4}$ m/s

Hope this answer is helpful.