A locomotive of mass m starts moving so that its velocity varies according to the law `v=asqrts`, where a is a constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the locomotive during the first t seconds after the beginning of motion.

Options

A. `8 " m "a^(4)t^(2)`
B. `4 " m "a^(4)t^(2)`
C. `(1)/(4) " m "a^(4)t^(2)`
D. `(1)/(8) " m "a^(4)t^(2)`

Answer

To solve the problem, we need to find the total work performed by all the forces acting on a locomotive of mass \( m \) during the first \( t \) seconds after it starts moving, with the velocity given by the equation \( v = a \sqrt{s} \).

### Step 1: Relate velocity to distance
We start with the given equation for velocity:
\[ v = a \sqrt{s} \]

### Step 2: Find acceleration
Acceleration \( a \) can be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can express this as:
\[ a = \frac{dv}{ds} \cdot \frac{ds}{dt} \]
Since \( \frac{ds}{dt} = v \), we can rewrite the acceleration as:
\[ a = \frac{dv}{ds} \cdot v \]

### Step 3: Differentiate the velocity equation
Now, we differentiate \( v = a \sqrt{s} \) with respect to \( s \):
\[ \frac{dv}{ds} = \frac{a}{2\sqrt{s}} \]
Substituting this into our expression for acceleration, we have:
\[ a = \frac{a}{2\sqrt{s}} \cdot v \]

### Step 4: Substitute \( v \) in terms of \( s \)
Substituting \( v = a \sqrt{s} \) into the acceleration equation gives:
\[ a = \frac{a}{2\sqrt{s}} \cdot a \sqrt{s} = \frac{a^2}{2} \]
Thus, the acceleration \( a' \) is:
\[ a' = \frac{a^2}{2} \]

### Step 5: Calculate the distance covered in time \( t \)
Using the kinematic equation for distance covered when starting from rest:
\[ s = ut + \frac{1}{2} a' t^2 \]
Since the initial velocity \( u = 0 \):
\[ s = \frac{1}{2} \left(\frac{a^2}{2}\right) t^2 = \frac{a^2 t^2}{4} \]

### Step 6: Calculate the work done
The work done \( W \) is given by:
\[ W = F \cdot s \]
Where \( F \) is the force, given by Newton's second law \( F = m a' \):
\[ F = m \cdot \frac{a^2}{2} \]
Substituting for \( F \) and \( s \):
\[ W = \left(m \cdot \frac{a^2}{2}\right) \cdot \left(\frac{a^2 t^2}{4}\right) \]
Simplifying this gives:
\[ W = \frac{m a^4 t^2}{8} \]

### Final Answer:
The total work performed by all the forces acting on the locomotive during the first \( t \) seconds is:
\[ W = \frac{m a^4 t^2}{8} \]

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