Let's tackle question 57 step-by-step:
To find the minimum time required for a vehicle of mass [tex]\( m \)[/tex] being driven by an engine of power [tex]\( P \)[/tex] to accelerate from rest, we analyze the kinematics and the relationship between power, force, and acceleration.
1. Understand Power and Force Relationships:
- Power ([tex]\( P \)[/tex]) is the rate at which work is done.
- Work done ([tex]\( W \)[/tex]) is given by [tex]\( W = F \cdot d \)[/tex], where [tex]\( F \)[/tex] is the force and [tex]\( d \)[/tex] is the distance.
- We know that power is also given by the product of force and velocity: [tex]\( P = F \cdot v \)[/tex].
2. Relate Force and Acceleration:
- Newton's second law states [tex]\( F = m \cdot a \)[/tex], where [tex]\( a \)[/tex] is the acceleration.
- Substituting [tex]\( F = m \cdot a \)[/tex] in the power equation [tex]\( P = F \cdot v \)[/tex], we get:
[tex]\[
P = m \cdot a \cdot v.
\][/tex]
- Rearranging for acceleration, [tex]\( a \)[/tex]:
[tex]\[
a = \frac{P}{m \cdot v}.
\][/tex]
3. Set Up the Differential Equation:
- The acceleration [tex]\( a \)[/tex] is the time derivative of velocity ([tex]\( v \)[/tex]):
[tex]\[
a = \frac{dv}{dt}.
\][/tex]
- Thus:
[tex]\[
\frac{dv}{dt} = \frac{P}{m \cdot v}.
\][/tex]
- Rearrange and separate variables:
[tex]\[
v \, dv = \frac{P}{m} \, dt.
\][/tex]
4. Integrate to Find Time:
- Integrate both sides to find the velocity as a function of time:
[tex]\[
\int_{0}^{v} v \, dv = \frac{P}{m} \int_{0}^{t} dt.
\][/tex]
- The integrals yield:
[tex]\[
\left[ \frac{v^2}{2} \right]_{0}^{v} = \frac{P}{m} \left[ t \right]_{0}^{t}.
\][/tex]
- Simplify the result:
[tex]\[
\frac{v^2}{2} = \frac{P}{m} t.
\][/tex]
- Solving for time [tex]\( t \)[/tex]:
[tex]\[
t = \frac{m v^2}{2P}.
\][/tex]
Therefore, the minimum time required for a vehicle of mass [tex]\( m \)[/tex] driven by an engine of power [tex]\( P \)[/tex] to reach a velocity [tex]\( v \)[/tex] from rest is:
[tex]\[
t = \frac{m v^2}{2P}.
\][/tex]
