Question

A particle moving along a straight line has a velocity v m/s , when it cleared a distance of x m. These two are connected by the relation $\rm\:v=\sqrt{49+x}$. When its velocity is 1 m/s , its acceleration is...

A) 2
B) 7
C) 1
D) 0.5

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Answer 1

Answer:

Given:

Velocity vs Position relationship is as follows ;

$v = \sqrt{49 + x}$

To find:

Acceleration of the body when the velocity is 1 m/s.

Concept:

Acceleration can be expressed in the form of velocity and displacement in the following relationship (using chain rule).

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \red{ \sf{a = v \dfrac{dv}{dx} }}}$

Calculation:

Trying to find the value of dv/dx :

$v = \sqrt{49 + x}$

$= > \dfrac{dv}{dx} = \dfrac{1}{2 \sqrt{49 + x} }$

Now , acceleration is:

$\therefore \: a = v \dfrac{dv}{dx}$

$= > a = (\sqrt{49 + x} ) \bigg \{ \dfrac{1}{(2 \sqrt{49 + x} )} \bigg \}$

$= > a = \dfrac{1}{2}$

$= > a = 0.5 \: m {s}^{ - 2}$

So final answer :

$\boxed{ \green{ \sf{ \bold{ \huge{ a = 0.5 \: m {s}^{ - 2} }}}}}$

Answer 2

Explanation:

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