Kinematics and Dynamics

Kinematics describes motion; dynamics explains it in terms of forces. The first half of this topic needs calculus and the constant-acceleration formulae; the second needs $F = ma$ and a clear free-body diagram.

Take $g = 9.81\ \text{m s}^{-2}$.

Displacement, velocity and acceleration

$$v = \frac{ds}{dt} \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$

and going the other way,

$$s = \int v\,dt \qquad v = \int a\,dt$$

A particle has $v = 3t^2 - 12t + 9$ m s$^{-1}$.

$$a = 6t - 12$$

At rest when $v = 0$: $3(t-1)(t-3) = 0$, so $t = 1$ and $t = 3$ seconds.

With $s = 0$ at $t = 0$, integrating gives $s = t^3 - 6t^2 + 9t$.

Constant acceleration

When $a$ is constant, five quantities are linked by five equations:

$$v = u + at \qquad s = ut + \tfrac12 at^2 \qquad v^2 = u^2 + 2as$$ $$s = \frac{(u+v)}{2}t \qquad s = vt - \tfrac12 at^2$$

Each omits one variable. Identify the three you know and the one you want, and the right equation follows.

A car accelerates from rest at $2.5\ \text{m s}^{-2}$ for 8 s.

$$v = 0 + 2.5(8) = 20\ \text{m s}^{-1} \qquad s = 0 + \tfrac12(2.5)(64) = 80\ \text{m}$$

Vertical motion under gravity

Take one direction as positive and stay with it. Choosing upwards positive gives $a = -9.81$ throughout, including on the way down.

A ball is thrown vertically upwards at $20\ \text{m s}^{-1}$.

At the highest point $v = 0$:

$$0 = 20^2 + 2(-9.81)s \quad\Longrightarrow\quad s = \frac{400}{19.62} = 20.4\ \text{m}$$

$$t = \frac{0 - 20}{-9.81} = 2.04\ \text{s to the top, so } 4.08\ \text{s in the air}$$

Velocity–time graphs

• Gradient = acceleration
• Area under the graph = displacement
• Area below the axis counts as negative displacement

A trapezium is the usual shape: constant acceleration, a steady stage, then constant deceleration. Its area, $\tfrac12(a+b)h$, gives the total distance in one step.

Newton's laws

1. A body stays at rest or moves with constant velocity unless a resultant force acts.
2. $F = ma$, with $F$ the resultant force, in the direction of the acceleration.
3. Action and reaction are equal and opposite, and act on different bodies.

A car of mass 1200 kg has a driving force of 3000 N against a resistance of 800 N.

$$F = 3000 - 800 = 2200\ \text{N} \qquad a = \frac{2200}{1200} = 1.83\ \text{m s}^{-2}$$

Connected particles

Two masses joined by a light inextensible string over a smooth pulley share one acceleration and one tension. Write $F = ma$ for each, taking the direction of motion as positive for each.

Masses of 5 kg and 3 kg.

$$5g - T = 5a \qquad T - 3g = 3a$$

Adding eliminates $T$:

$$2g = 8a \quad\Longrightarrow\quad a = \frac{2(9.81)}{8} = 2.45\ \text{m s}^{-2}$$

$$T = 3(g + a) = 3(12.26) = 36.8\ \text{N}$$

Momentum and impulse

$$\text{momentum} = mv \qquad \text{impulse} = Ft = mv - mu$$

In any collision, total momentum is conserved:

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

A 2 kg body at $5\ \text{m s}^{-1}$ strikes a stationary 3 kg body and they move off together.

$$2(5) + 3(0) = 5v \quad\Longrightarrow\quad v = 2\ \text{m s}^{-1}$$

The impulse on the 3 kg body is $3(2) - 0 = 6\ \text{N s}$, and on the 2 kg body $2(2) - 2(5) = -6\ \text{N s}$ — equal and opposite, as Newton's third law requires.