If velocity v of particle moving in straight line is related with distance travelled S as v=2(1+S)^1/2. What will the acceleration of the particle?

The problem I have here is units.

velocity is distance/time, and your equation is sqrt distance. Hmmmm.

Assuming somehow that is fixed in magic land, then

acceleration= dv/dt=
(1+s)^-1/2 * ds/dt but ds/dt is V so

acceleration= 2 (1+s)^0=2

To find the acceleration of the particle, we need to differentiate the given equation with respect to time (t). The derivative of velocity with respect to time gives us the acceleration.

Given:
v = 2(1 + S)^(1/2)

Differentiating both sides of the equation with respect to time (t):

dv/dt = d/dt [2(1 + S)^(1/2)]

To differentiate the right-hand side of the equation, we can use the chain rule of differentiation. Let's denote (1 + S) as u.

du/dt = d(1 + S)/dS * dS/dt

Since we are given the relation v = 2(1 + S)^(1/2), we can solve for S as follows:

v = 2(1 + S)^(1/2)
v/2 = (1 + S)^(1/2)
(v/2)^2 = 1 + S
S = (v^2/4) - 1

Now, let's differentiate S with respect to t:

dS/dt = d[((v^2)/4) - 1]/dt
= (1/4) * d(v^2)/dt
= (1/4) * 2v * dv/dt
= (v/2) * dv/dt

Substituting the value of dS/dt back into our original equation:

dv/dt = d/dt [2(1 + S)^(1/2)]
= d/dt [2u^(1/2)]
= (1/2) * d(2u)^(1/2)/d(t)
= (1/2) * (du/dt)/(u^(1/2))
= (1/2) * (v/2) * dv/dt * (2(1 + S)^(1/2))^(-1/2)
= (v/4) * dv/dt * (1 + S)^(-1/2)

Finally, we have the expression for the acceleration:

a = dv/dt
= (v/4) * dv/dt * (1 + S)^(-1/2)

Substituting the value of dS/dt we found earlier:

a = (v/4) * (v/2) * dv/dt * (1 + S)^(-1/2)
= (v^2/8) * (dv/dt) * (1 + S)^(-1/2)

Therefore, the acceleration of the particle is (v^2/8) * (dv/dt) * (1 + S)^(-1/2).

To find the acceleration of the particle, we need to differentiate the velocity equation with respect to time.

Given:
Velocity, v = 2(1 + S)^(1/2)

Differentiating with respect to time (t) using the chain rule:
dv/dt = d/dt [2(1 + S)^(1/2)]

To evaluate the derivative, we need to use the chain rule, which states that if we have a composition of two or more functions, then their derivatives can be found by differentiating each function separately and multiplying them together.

Let's break down the differentiation:

Let u = 1 + S
Using the chain rule, we have:
dv/dt = d/dt[2u^(1/2)]

Now, let's differentiate the function 2u^(1/2) with respect to u:
du/dt = 1/2 * 2u^(-1/2) * du/dt
Simplifying further:
du/dt = u^(-1/2) * du/dt

Substituting back u = 1 + S:
du/dt = (1 + S)^(-1/2) * dS/dt

Now, we need to find dS/dt, which is the rate of change of distance traveled with respect to time. This represents the velocity, v.

Therefore:
dS/dt = v

Substituting back into the previous equation:
du/dt = (1 + S)^(-1/2) * v

Now, let's substitute all the derivatives back into the original equation:
dv/dt = du/dt * dS/dt

Substituting:
dv/dt = (1 + S)^(-1/2) * v * v

Simplifying:
dv/dt = v^2 / (1 + S)^(1/2)

This expression represents the acceleration, a, of the particle.
Therefore, the acceleration is given by:
a = v^2 / (1 + S)^(1/2)

Note: Keep in mind that this expression for acceleration depends on the distance traveled, S, and the velocity, v, at that specific distance.