Question

A particle moving along a straight line has a velocity v m/s , when it cleared a distance of x meters. These two are connected by the relation v= root 49+x find the acceleration of the particle in m/s square when its velocity is 1 m/s

Answer 1

$\huge\bold{HELLO\:MATE}$

follow these steps→

$\huge\bold{1→}$equation of v is given.

$\huge\bold{2→}$ to derive the equation of acceleration, differentiate the equation of v with time.

$\huge\bold{3→}$ the final equation will give the relationship between acceleration and velocity.

$\huge\bold{4→}$ put the value of velocity in the derived equation and hence value of acceleration can be found.

$\huge\boxed{solution→}$

$v \: = \sqrt{49 + y}$

acceleration (a) =

$\huge \bold{ \frac{dv}{dt} = \frac{1}{2 \sqrt{49 + y} } }$

$\huge {\frac{1}{2 \sqrt{49 + y} } \times \frac{dy}{dt} }$

${\frac{dy}{dt} = \: velocity = 1m {s}^{ - 1} }$

$\huge \bold{ a = \frac{dv}{dt} = \frac{1}{2 \sqrt{49 + y} }}$

❤️hence proved ❤️

Answer 2

Answer:

Explanation:

Solution,

Here, we have

v = √49 + x

Squaring both sides, we get

v² = 49 + x

Differentiating both the sides w.r.t, we get

⇒ 2v(dv/dt) = dx/dt

Here, we know that

v = dx/dt

Then,

Putting the value, we get

⇒ 2v(dv/dt) = v    

⇒ dv/dt = 1/2

⇒ dv/dt = 0.5

Acceleration, a = dv/dt = 0.5 m/s².

Hence, the accelerations is 0.5 m/s².