Kinematics with calculus for variable acceleration, vectors in kinematics, and projectile motion resolved into horizontal and vertical components.

What this dot point is asking

WJEC wants you to use calculus for motion with variable acceleration (differentiating displacement to velocity to acceleration, and integrating back), to handle vectors in kinematics, and to analyse projectile motion by resolving the velocity into independent horizontal and vertical components. This extends the constant-acceleration kinematics of AS Unit 2 to the general case.

The answer

Kinematics with calculus

When acceleration is not constant, the suvat equations do not apply, so use calculus instead.

Differentiation moves down the chain (displacement to velocity to acceleration); integration moves back up, with a constant fixed by an initial condition.

Vectors in kinematics

Displacement, velocity and acceleration can be written as vectors, for example $\mathbf{r} = (t^2)\mathbf{i} + (3t)\mathbf{j}$. Differentiate each component separately to get the velocity and acceleration vectors. The speed is the magnitude of the velocity vector, $\sqrt{v_x^2 + v_y^2}$. For example, if $\mathbf{v} = (4t)\mathbf{i} + (3)\mathbf{j}$, then at $t = 1$ the velocity is $4\mathbf{i} + 3\mathbf{j}$ and the speed is $\sqrt{16 + 9} = 5\,\text{m s}^{-1}$. The acceleration vector here is the derivative $\mathbf{a} = 4\mathbf{i}$, constant in the $\mathbf{i}$ direction.

Projectile motion

A projectile moves under gravity alone. Resolve the launch velocity into $u_x = u\cos\theta$ and $u_y = u\sin\theta$.

Examples in context

Example 1. Maximum speed from acceleration. A particle has $v = 6t - t^2$. Its acceleration is $a = \dfrac{dv}{dt} = 6 - 2t$, zero at $t = 3$, where the velocity reaches its maximum of $v = 18 - 9 = 9\,\text{m s}^{-1}$. Setting the acceleration to zero finds the turning point of the velocity.

Example 2. Range on level ground. For a projectile launched at speed $u$ and angle $\theta$ on level ground, the range is $R = \dfrac{u^2 \sin 2\theta}{g}$, which is greatest at $\theta = 45^{\circ}$ because $\sin 2\theta$ peaks there. Resolving and using the time of flight derives this standard result.

Try this

Q1. A particle has displacement $s = 2t^2 - t$. Find its velocity at $t = 3$. [2 marks]

• Cue. $v = \dfrac{ds}{dt} = 4t - 1 = 11\,\text{m s}^{-1}$ at $t = 3$.

Q2. A projectile is launched at $30\,\text{m s}^{-1}$ at $60^{\circ}$. Find the vertical component of the initial velocity. [2 marks]

• Cue. $u_y = 30\sin 60^{\circ} = 25.98\,\text{m s}^{-1}$.

Q3. A particle has acceleration $a = 6t$. Given $v = 2$ at $t = 0$, find $v$ as a function of $t$. [3 marks]

• Cue. $v = \int 6t\,dt = 3t^2 + c$; the condition gives $c = 2$, so $v = 3t^2 + 2$.