A car of mass `m` starta from rest and accelerates so that the instyantaneous power delivered to the car has a constant magnitude `P_(0)`. The instaneous velocity of this car is proportional to

Options

A. `t^(1//2)`
B. `t^(-1//2)`
C. `t//sqrtm`
D. `t^(2)P_(0)`

Explanation

To solve the problem, we need to analyze the relationship between power, force, and velocity for the car that starts from rest and accelerates under the influence of a constant power \( P_0 \).

### Step-by-Step Solution:

1. **Understanding Power**:
Power \( P \) is defined as the rate at which work is done or energy is transferred. Mathematically, it can be expressed as:
\[ P = F \cdot v \]
where \( F \) is the force applied to the car and \( v \) is its instantaneous velocity.

2. **Relating Force to Acceleration**:
According to Newton's second law, the force acting on the car can be expressed as:
\[ F = m \cdot a \]
where \( m \) is the mass of the car and \( a \) is its acceleration.

3. **Substituting Force in Power Equation**:
By substituting the expression for force into the power equation, we get:
\[ P = m \cdot a \cdot v \]

4. **Acceleration in Terms of Velocity**:
Since the car starts from rest, we can express acceleration \( a \) in terms of velocity \( v \) and time \( t \):
\[ a = \frac{dv}{dt} \]
Therefore, we can rewrite the power equation as:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]

5. **Rearranging the Power Equation**:
Rearranging gives us:
\[ P = m \cdot v \cdot \frac{dv}{dt} \]

6. **Separating Variables**:
We can separate the variables to integrate:
\[ \frac{P}{m} = v \cdot \frac{dv}{dt} \]
This can be rewritten as:
\[ \frac{P}{m} dt = v \, dv \]

7. **Integrating Both Sides**:
Integrating both sides, we have:
\[ \int 0^{t} \frac{P}{m} dt = \int 0^{v} v \, dv \]
The left side becomes:
\[ \frac{P}{m} t \]
The right side becomes:
\[ \frac{1}{2} v^2 \]
Thus, we have:
\[ \frac{P}{m} t = \frac{1}{2} v^2 \]

8. **Solving for Velocity**:
Rearranging the equation for \( v^2 \):
\[ v^2 = \frac{2P}{m} t \]
Taking the square root gives:
\[ v = \sqrt{\frac{2P}{m} t} \]

9. **Identifying Proportionality**:
From the equation \( v = \sqrt{\frac{2P}{m}} \sqrt{t} \), we can see that the instantaneous velocity \( v \) is proportional to \( \sqrt{t} \).

### Final Answer:
The instantaneous velocity of the car is proportional to \( \sqrt{t} \).