A cycle rides through a puddle. Drops of muddy water leave the rim of the wheel. They do not fly outward along the spokes. Each drop leaves along the tangent to the rim at the point where it lets go, in the direction the rim was moving at that instant. A stone released from a sling behaves the same way.
Average velocity is a vector
Average velocity in a plane is defined exactly as on a line. It is the displacement divided by the time interval. Because the displacement is a vector, the average velocity is a vector too. Its direction is the direction of the displacement.
\[ \vec v_{\text{avg}} = \dfrac{\Delta\vec r}{\Delta t} = \dfrac{\Delta x}{\Delta t}\,\hat i + \dfrac{\Delta y}{\Delta t}\,\hat j \]
The division by \(\Delta t\) is done component by component. Each component of the average velocity is the average velocity along that axis.
Instantaneous velocity points along the tangent
Shrink the time interval towards zero, as in the last chapter. The displacement \(\Delta\vec r\) becomes a very short arrow along the path. A very short piece of a curve is a straight piece, and its direction is the direction of the tangent to the curve at that point. So the instantaneous velocity is always directed along the tangent to the path.
\[ \vec v = \lim_{\Delta t \to 0}\dfrac{\Delta\vec r}{\Delta t} = \dfrac{d\vec r}{dt} = \dfrac{dx}{dt}\,\hat i + \dfrac{dy}{dt}\,\hat j = v_x\,\hat i + v_y\,\hat j \]
The magnitude of \(\vec v\) is the speed, \(|\vec v| = \sqrt{v_x^2 + v_y^2}\). The components \(v_x\) and \(v_y\) are the velocities along the two axes, each obtained from its own coordinate by differentiation.
The instantaneous velocity of a body is directed along the tangent to its path at the position of the body, in the sense of motion. Its magnitude is the speed.
A particle moves so that \(x = 3t\) and \(y = 4t - 5t^2\), in metres with \(t\) in seconds. Find the velocity vector and the speed at \(t = 0\) and at \(t = 1\,\text{s}\).
Differentiate each coordinate: \(v_x = dx/dt = 3\,\text{m/s}\), constant, and \(v_y = dy/dt = 4 - 10t\).
At \(t = 0\): \(\vec v = 3\hat i + 4\hat j\), speed \(= 5\,\text{m/s}\).
At \(t = 1\,\text{s}\): \(\vec v = 3\hat i - 6\hat j\), speed \(= \sqrt{9 + 36} = \sqrt{45} \approx 6.7\,\text{m/s}\).
The \(x\) component never changed; the \(y\) component reversed its sign, so the particle first rose and then fell.
A stone is whirled in a horizontal circle on a string. The string breaks. In which direction does the stone move at the instant of breaking?
Acceleration in a plane
Acceleration is the rate of change of velocity. In a plane, velocity is a vector, so acceleration is the rate of change of a vector. A vector can change in two ways. Its magnitude can change, and its direction can change. Acceleration is present in either case.
\[ \vec a = \dfrac{d\vec v}{dt} = \dfrac{dv_x}{dt}\,\hat i + \dfrac{dv_y}{dt}\,\hat j = a_x\,\hat i + a_y\,\hat j \]
Each component of the acceleration is the derivative of the corresponding component of the velocity.
Look at Fig 2.2. The part of \(\vec a\) along \(\vec v\) makes the body speed up or slow down. The part of \(\vec a\) perpendicular to \(\vec v\) turns the velocity without changing its magnitude. A body moving at constant speed around a curve has an acceleration that is entirely perpendicular to its velocity. A body moving along a straight line has an acceleration entirely along its velocity, as in the last chapter.
The component of the acceleration along the velocity changes the speed of the body. The component of the acceleration perpendicular to the velocity changes the direction of motion without changing the speed. A body with constant speed on a curved path still has an acceleration, directed perpendicular to its velocity.
Constant speed does not mean zero acceleration. A car rounding a bend at a steady \(40\,\text{km/h}\) is accelerating, because the direction of its velocity is changing. The passengers feel this acceleration as a push towards the outside of the bend.
A particle starts at the origin with velocity \(\vec u = 3\hat i\,\text{m/s}\) and moves with constant acceleration \(\vec a = -2\hat j\,\text{m/s}^2\). Find its velocity and position at \(t = 2\,\text{s}\).
The equations of motion hold for each component separately, because the acceleration is constant.
Along \(x\): \(a_x = 0\), so \(v_x = 3\,\text{m/s}\) and \(x = 3t = 6\,\text{m}\).
Along \(y\): \(a_y = -2\,\text{m/s}^2\), \(u_y = 0\), so \(v_y = -4\,\text{m/s}\) and \(y = \tfrac12(-2)(4) = -4\,\text{m}\).
At \(t = 2\,\text{s}\): \(\vec v = 3\hat i - 4\hat j\,\text{m/s}\), speed \(5\,\text{m/s}\), position \((6, -4)\,\text{m}\).
The velocity turned downward because the acceleration was perpendicular to the initial velocity, and the speed rose from \(3\) to \(5\,\text{m/s}\) because the acceleration is no longer perpendicular to the velocity once the velocity has turned.
A body moves at constant speed along a curved path. Which statement is correct?
A particle moves in a plane with \(x = 2t\) and \(y = t^2\), in SI units. Which statements are true?
The equations of motion as vectors
Suppose the acceleration \(\vec a\) is constant, in magnitude and in direction. Then each of its components \(a_x\) and \(a_y\) is constant too. Along the \(x\) axis the motion is uniformly accelerated, so the equations of the last chapter hold for \(x\). The same is true along the \(y\) axis. Write each pair of equations together, one component beside the other, and two vector equations appear.
\[ \vec v = \vec u + \vec a\,t, \qquad \vec r = \vec r_0 + \vec u\,t + \tfrac12\,\vec a\,t^2 \]
Here \(\vec r_0\) is the position vector at \(t = 0\) and \(\vec u\) is the velocity at \(t = 0\). Each vector equation is a short way of writing two ordinary equations, one for each axis.
Read the second equation as a sum of arrows. The term \(\vec u\,t\) is where the body would go with no acceleration at all, along a straight line in the direction of \(\vec u\). The term \(\tfrac12\,\vec a\,t^2\) is the extra displacement that the acceleration adds, always in the direction of \(\vec a\). So the path bends away from the straight line \(\vec u\,t\), towards the direction of \(\vec a\), by an amount that grows as \(t^2\).
The third equation of motion, \(v^2 = u^2 + 2as\), has no simple vector form here. Use it one component at a time: \(v_x^2 = u_x^2 + 2a_x\,\Delta x\) and \(v_y^2 = u_y^2 + 2a_y\,\Delta y\). Adding the squares of speeds and displacements directly gives wrong answers.
A third axis
Nothing in this argument needed exactly two axes. A body moving in space, such as a drone or a ball kicked across a field, needs three coordinates. A third axis \(z\) is added, perpendicular to both \(x\) and \(y\), and its unit vector is written \(\hat k\). The position vector becomes \(\vec r = x\,\hat i + y\,\hat j + z\,\hat k\), and every equation above gains a \(\hat k\) component. The magnitude of a vector in space is \(\sqrt{x^2 + y^2 + z^2}\). Most motions in this book stay in one plane, so the \(\hat k\) component is zero there.
A particle is at \(\vec r_0 = 2\hat i + \hat j\,\text{m}\) at \(t = 0\), moving with \(\vec u = 4\hat i + 6\hat j\,\text{m/s}\). Its acceleration is constant, \(\vec a = \hat i - 2\hat j\,\text{m/s}^2\). Find its velocity and position at \(t = 2\,\text{s}\).
Velocity: \(\vec v = \vec u + \vec a\,t = (4 + 2)\hat i + (6 - 4)\hat j = 6\hat i + 2\hat j\,\text{m/s}\). The speed is \(\sqrt{36 + 4} = \sqrt{40} \approx 6.3\,\text{m/s}\).
Position: \(\vec r = \vec r_0 + \vec u\,t + \tfrac12\vec a\,t^2 = (2 + 8 + 2)\hat i + (1 + 12 - 4)\hat j = 12\hat i + 9\hat j\,\text{m}\).
The particle is then \(\sqrt{144 + 81} = 15\,\text{m}\) from the origin. Each component was worked out with its own one-dimensional equation, and the vector form only kept the two side by side.
A particle has initial velocity \(\vec u = 3\hat i\,\text{m/s}\) and a constant acceleration \(\vec a = 2\hat j\,\text{m/s}^2\). What is its speed at \(t = 2\,\text{s}\), in m/s?
A drone starts from rest at the origin. Its acceleration is constant, \(\vec a = \hat i + 2\hat j + 2\hat k\,\text{m/s}^2\). How far from the origin is the drone after \(2\,\text{s}\)?
On a bicycle without mudguards, mud from the rear wheel lands on the rider's back. A drop leaves the rim along the tangent, in the direction the rim is moving at that point. At the top of the wheel the rim moves forward, so drops leaving there fly forward. At the back of the wheel the rim moves upward, so drops leaving there fly upward, and the forward motion of the cycle carries them onto the rider. A mudguard covers exactly the points of the rim whose tangent points at the rider.
- Drawing the velocity of a body on a curve along the radius or towards the centre.Velocity is always along the tangent, in the sense of motion.
- Saying that constant speed means zero acceleration.Ask whether the direction of the velocity changes. On a curve it does, and the acceleration is perpendicular to the velocity.
- Differentiating the magnitude of the position instead of the coordinates.Differentiate \(x(t)\) and \(y(t)\) separately. The speed is then \(\sqrt{v_x^2 + v_y^2}\), which is not \(d|\vec r|/dt\).
- Adding the components of a vector as plain numbers to get its magnitude.Components are at right angles. The magnitude is \(\sqrt{x^2 + y^2 + z^2}\), never \(x + y + z\).