No history yet

Variable Acceleration Analysis

Motion Beyond Straight Lines

In introductory physics, we often work with constant acceleration. A car speeds up steadily, a ball falls under gravity—the change in velocity is uniform. This simplifies the maths, but the real world is rarely so neat. A rocket's acceleration changes as it burns fuel and gets lighter. A car's braking is not perfectly constant. To describe this kind of complex movement, we need to think about motion not as a set of fixed numbers, but as a continuous function of time.

This is where calculus comes in. It's the language of change. If we know an object's position, ss, at any given time, tt, we can use differentiation to find its instantaneous velocity. Likewise, we can differentiate the velocity function to find its instantaneous acceleration. This allows us to pinpoint the exact rate of change at any moment.

v(t)=dsdtanda(t)=dvdt=d2sdt2v(t) = \frac{ds}{dt} \quad \text{and} \quad a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Consider a particle whose displacement in metres is given by the function s(t)=2t3−15t2+36t−10s(t) = 2t^3 - 15t^2 + 36t - 10, where tt is in seconds. We can find its velocity and acceleration at any time by differentiating this function.

# Differentiating to find velocity
v(t) = d/dt (2t³ - 15t² + 36t - 10)
v(t) = 6t² - 30t + 36

# Differentiating again for acceleration
a(t) = d/dt (6t² - 30t + 36)
a(t) = 12t - 30

With these derived functions, we can answer detailed questions. For instance, when is the particle momentarily at rest? That happens when v(t)=0v(t) = 0. Solving 6t2−30t+36=06t^2 - 30t + 36 = 0 gives us t=2t = 2 seconds and t=3t = 3 seconds. The acceleration isn't constant; it changes linearly with time. At t=2t=2 s, the acceleration is a(2)=12(2)−30=−6 m/s2a(2) = 12(2) - 30 = -6 \, \text{m/s}^2.

Rebuilding Motion with Integration

Just as differentiation breaks motion down, integration builds it back up. If we know the acceleration of an object over time, we can integrate to find its change in velocity. Integrating the velocity function gives us its change in position, or displacement. This process is the cornerstone of solving problems in , like that of a rocket losing mass as it flies.

Δv=vf−vi=∫titfa(t) dtΔs=sf−si=∫titfv(t) dt\Delta v = v_f - v_i = \int_{t_i}^{t_f} a(t) \, dt \\ \Delta s = s_f - s_i = \int_{t_i}^{t_f} v(t) \, dt

This relationship is powerfully visualised in graphs. For non-linear motion, the area under the acceleration-time graph gives the change in velocity, and the area under the velocity-time graph gives the displacement. While for straight lines this area is a simple rectangle or triangle, for curved graphs, integration is the only precise tool to calculate that area.

When we perform indefinite integration, we get a family of possible solutions that differ by a , C. To find the specific equation of motion for an object, we need more information: its initial conditions. For example, knowing the velocity at t=0t=0 allows us to solve for CC in the velocity equation. Knowing the initial position lets us solve for the constant in the displacement equation. This anchors the general solution to a specific, real-world scenario.

The fundamental kinematic equations you learned for constant acceleration are actually just special cases derived by integrating a constant acceleration, aa, twice with respect to time.

Quiz Questions 1/5

Why are the standard kinematic equations (like s=ut+12at2s = ut + \frac{1}{2}at^2) often insufficient for describing real-world motion, necessitating the use of calculus?

Quiz Questions 2/5

A particle's position is described by the function s(t)=2t3−15t2+36t−10s(t) = 2t^3 - 15t^2 + 36t - 10, where ss is in metres and tt is in seconds. At which of the following times is the particle momentarily at rest?

By mastering the application of differentiation and integration, you can analyze any motion, no matter how its acceleration varies. This is the true power of applying calculus to mechanics.