Question

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P_0$. The instantaneous velocity of this car is proportional to

• A. $t^2 P_0$
• B. $t^{1/2}$
• C. $t^{-1/2}$
• D. $\frac{1}{\sqrt m}$

Answer 1

The Correct Option is B

To solve this problem, we need to determine how the instantaneous velocity of the car is related to time, given that the instantaneous power delivered to the car has a constant magnitude \( P_0 \).

Step 1: Understand the relation between power, force, and velocity.

The power \( P \) delivered to an object is given by the equation:

P = Fv

Where \( F \) is the force applied on the object and \( v \) is the velocity of the object.

Step 2: Relate power to velocity and time.

Since \( P_0 \) is constant, we have:

P_0 = Fv

If we assume the car's acceleration \( a \) is due to the force \( F \), then by Newton's second law, \( F = ma \). Therefore:

P_0 = ma \cdot v

Substitute acceleration \( a = \frac{dv}{dt} \), we have:

P_0 = m \frac{dv}{dt} \cdot v

Step 3: Solve the differential equation.

Rearranging the terms, we have:

m v \frac{dv}{dt} = P_0

Integrate with respect to time:

Separate the variables:

v \, dv = \frac{P_0}{m} \, dt

Integrate both sides:

\int v \, dv = \int \frac{P_0}{m} \, dt

This gives:

\frac{v^2}{2} = \frac{P_0}{m} t + C

Since the car starts from rest at \( t = 0 \), the constant \( C = 0 \).

Step 4: Derive the relationship between velocity and time.

Now, solve for \( v \):

v^2 = \frac{2P_0}{m} t

Therefore, the velocity \( v \) is given by:

v = \sqrt{\frac{2P_0 t}{m}}

Notice that v \propto \sqrt{t}

Conclusion:

The instantaneous velocity of the car is proportional to t^{1/2}. Therefore, the correct answer is \( t^{1/2} \).