The engine of a motorbike produces a constant power \(P\). The bike starts at rest and drives in a straight line. We neglect effects of friction and air resistance. (a) Find the velocity of the bike as a function of time? (b) Find the acceleration and show that it is not a constant. (c) Find the position, \(x(t)\), for the bike as a function of time.

In physics, power is the rate at which work is done or energy is transferred over time. For a motorbike moving in a straight line with a constant power output and negligible friction or air resistance, this relationship can be succinctly described by the equation: \[ P = F \times v \] Here, \(P\) represents constant power, \(F\) is the force applied by the motorbike engine, and \(v\) is the velocity. This equation highlights that power is the product of force and velocity. To find the force in terms of power and velocity, we rearrange this equation to: \[ F = \frac{P}{v} \] This foundational relationship helps us understand how power influences velocity in motion scenarios such as the one provided.

Newton's Second Law of Motion provides a pivotal connection between force, mass, and acceleration. It states: \[ F = m \times a \] Where \(F\) is force, \(m\) is mass, and \(a\) is acceleration. When combined with the previously derived expression for force\( F = \frac{P}{v} \), we get: \[ m \times a = \frac{P}{v} \] Thus, acceleration \(a\) can be represented as: \[ a = \frac{P}{m \times v} \] This relationship shows how acceleration depends directly on power and inversely on mass and velocity.

To determine how velocity changes over time for a motorbike under constant power, we start with the equation linking acceleration to power and velocity: \[ a = \frac{P}{m \times v} \] Since acceleration is the time derivative of velocity \( \frac{dv}{dt} \), we can write: \[ \frac{dv}{dt} = \frac{P}{m \times v} \] Separating variables and integrating, we get: \[ v\ dv = \frac{P}{m}\ dt \] Integrating both sides, we find: \[ \frac{v^2}{2} = \frac{P}{m}t + C \] Using the initial condition that the bike starts from rest \((v = 0)\ at\ (t = 0)\), we find that the constant \(C = 0\). Thus, \[ v(t) = \sqrt{\frac{2P}{m}t} \] This equation gives us the velocity as a function of time.

From the derived velocity function \(v(t)\), \[ v(t) = \sqrt{\frac{2P}{m}t} \] we can determine acceleration by differentiating velocity with respect to time: \[ a(t) = \frac{d}{dt}\left( \sqrt{\frac{2P}{m}t} \right) = \frac{\frac{2P}{m}}{2}\left( \frac{1}{\sqrt{\frac{2P}{m}t}} \right) = \sqrt{\frac{P}{2m t}} \] This result shows that acceleration is not constant but depends on time as: \[ a(t) = \sqrt{\frac{P}{2m t}} \] This means that acceleration decreases over time as the motorbike moves.

Finally, to find the position \(x(t)\) as a function of time, we need to integrate the velocity function: \[ v(t) = \sqrt{\frac{2P}{m}t} \] Integrating with respect to time, we get: \[ x(t) = \int \sqrt{\frac{2P}{m}t}\ dt \] Using the substitution \(u = \sqrt{t}\), we have \( du = \frac{1}{2\sqrt{t}}\ dt\), thus \(dt = 2u\ du\). Substituting, we obtain: \[ x(t) = \int 2u^2\sqrt{\frac{2P}{m}}\ du \] Solving the integral, we find: \[ x(t) = \sqrt{\frac{2P}{m}}\cdot\frac{2}{3}\cdot\left( t^{3/2} \right) \] Thus, the position as a function of time is: \[ x(t) = \frac{2}{3}\sqrt{\frac{2P}{m}} t^{3/2} \] This represents how far the motorbike travels over a given time period under constant power.