Road-safety boards on the Mumbai-Pune Expressway warn drivers to keep a long gap at high speed. A car at \(20\,\text{m/s}\) that brakes at \(5\,\text{m/s}^2\) stops in \(40\,\text{m}\). The same car at \(40\,\text{m/s}\), braking just as hard, needs \(160\,\text{m}\). Twice the speed has made the stopping distance four times as long, not twice.
The square law
The stopping distance of a car with a constant deceleration \(a\) is \(s = u^2/2a\), where \(u\) is the speed when the brakes are applied. The distance is a constant, \(1/2a\), times the square of the speed. Such a relation is a square law, \(y = kx^2\). Multiply \(x\) by \(2\) and \(y\) is multiplied by \(2^2 = 4\). Multiply \(x\) by \(3\) and \(y\) is multiplied by \(9\).
If \(y = kx^2\) with \(k\) a constant, then multiplying \(x\) by any factor \(n\) multiplies \(y\) by \(n^2\). In ratios, \(y_2/y_1 = (x_2/x_1)^2\), and the constant \(k\) never needs to be known.
The square law is everywhere in physics. The kinetic energy \(\tfrac12 mv^2\) goes as the square of the speed. The energy of a stretched spring, \(\tfrac12 kx^2\), goes as the square of the extension. The distance fallen from rest, \(\tfrac12 gt^2\), goes as the square of the time. The graph of each of them is a parabola with its lowest point, the vertex, at the origin.
An auto-rickshaw triples its speed. What happens to its kinetic energy?
The steps of a falling stone
A stone is let go from the top of a tower \(80\,\text{m}\) high. With \(g = 10\,\text{m/s}^2\), the distance it has fallen after \(t\) seconds is \(5t^2\). After \(1\), \(2\), \(3\) and \(4\,\text{s}\) that is \(5\), \(20\), \(45\) and \(80\,\text{m}\). So the gaps covered in the four seconds, one after another, are \(5\), \(15\), \(25\) and \(35\,\text{m}\).
The gaps are in the ratio \(1 : 3 : 5 : 7\), the odd numbers. The reason is the square law. The distance up to the end of second \(n\) is \(5n^2\), and up to the end of second \(n - 1\) it is \(5(n - 1)^2\). The difference is \(5(2n - 1)\), five times an odd number. Galileo found this ratio by rolling balls down a gentle slope, long before calculus existed.
A stone is dropped from rest. Take \(g = 10\,\text{m/s}^2\). How far does it fall during the third second alone, from \(t = 2\,\text{s}\) to \(t = 3\,\text{s}\), in metres?
Every quadratic is a moved parabola
The general quadratic \(y = ax^2 + bx + c\) looks more complicated than \(y = kx^2\). It is not. Completing the square shows that it is the parabola \(y = ax^2\) moved sideways and up. The derivation below needs only the square \((x + p)^2 = x^2 + 2px + p^2\).
\[ y = a\left(x^2 + \dfrac{b}{a}x\right) + c \]
Why. take the factor \(a\) out of the two terms that contain \(x\); this needs \(a \ne 0\).
\[ x^2 + \dfrac{b}{a}x = \left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2} \]
Why. the square \((x + p)^2 = x^2 + 2px + p^2\) with \(p = b/2a\) has the same first two terms; subtract the extra \(p^2\).
\[ y = a\left(x + \dfrac{b}{2a}\right)^2 + c - \dfrac{b^2}{4a} \]
Why. put step 2 into step 1 and multiply the \(-b^2/4a^2\) by \(a\).
\[ x_v = -\dfrac{b}{2a}, \qquad y_v = c - \dfrac{b^2}{4a} \]
Read it. the squared bracket is zero at \(x = -b/2a\) and positive elsewhere, so \(y\) takes its extreme value \(y_v\) there.
For \(a > 0\) the squared term only adds, so the vertex is the lowest point and the parabola opens upward. For \(a < 0\) the squared term only subtracts, so the vertex is the highest point and the parabola opens downward. The squared bracket takes the same value at equal distances on either side of \(x_v\). So the parabola is symmetric about the vertical line \(x = x_v\), its axis of symmetry.
Where is the vertex of \(y = 2x^2 - 8x + 3\), and is it a highest or a lowest point?
Back to physics: the top of a thrown ball
The height of a ball thrown straight up is a quadratic in time with a negative coefficient of \(t^2\). So its graph is a parabola that opens downward, a dome, and the vertex is the top of the flight. Its roots, where the height is zero, are the moment of the throw and the moment of the catch. The lesson on the quadratic equation in the chapter Algebra for Physics finds such roots in general.
A ball leaves the hand at \(20\,\text{m/s}\) straight up. Take upward as positive and \(g = 10\,\text{m/s}^2\), so its height is \(y = 20t - 5t^2\). Find the time and the height of the top, and the time at which the ball returns to the hand.
Compare with \(y = at^2 + bt + c\): \(a = -5\), \(b = 20\), \(c = 0\). Since \(a < 0\), the vertex is the highest point.
\(t_v = -b/2a\) and \(y_v = c - b^2/4a\). The return is the second root of \(20t - 5t^2 = 0\).
\(t_v = -20/(2 \times (-5)) = 2\,\text{s}\) and \(y_v = 0 - 400/(-20) = 20\,\text{m}\). The roots of \(5t(4 - t) = 0\) are \(t = 0\) and \(t = 4\,\text{s}\).
The ball reaches the top, \(20\,\text{m}\), at \(2\,\text{s}\), and returns to the hand at \(4\,\text{s}\). These are the numbers of the physics chapter, whose v-t line crosses zero at \(2\,\text{s}\), exactly below the top of the dome.
The road-safety board is now easy to read. The stopping distance goes as the square of the speed. So a car at twice the speed needs four times the distance, and a car at three times the speed needs nine times. At \(20\,\text{m/s}\) the car needed \(40\,\text{m}\); at \(60\,\text{m/s}\) it would need \(360\,\text{m}\).
A ball is thrown at \(20\,\text{m/s}\) at \(53°\) to the horizontal, with \(\tan 53° = 4/3\) and \(\cos 53° = 0.6\). Its path is \(y = \tfrac43 x - x^2/28.8\), in metres. Find the greatest height of the path, in metres.
A spring stores \(0.8\,\text{J}\) when it is compressed by \(4\,\text{cm}\). How much energy does it store when it is compressed by \(10\,\text{cm}\), in joules?
A particle moving along the \(x\) axis has potential energy \(U = 5x^2 - 10x\), with \(U\) in joules and \(x\) in metres. What is the least value of \(U\), in joules?
A square law holds only while its conditions hold. The stopping distance \(u^2/2a\) needs the same deceleration all the way; worn brakes or a wet road change \(a\). A thrown ball follows a parabola only when air resistance is neglected. A root of the quadratic is a real moment only inside the domain. A negative time, before the throw, is not part of the flight.
The graphs of free fall in Graphical Understanding of Motion are parabolas in time. The path of every projectile in Motion in Two Dimensions is a parabola in space. The energy of a spring and the energy stored in a capacitor, \(q^2/2C\), follow square laws. The surface of a liquid in a spinning tank is a parabola, and so is a jet of water from a hole in a tank. The speed of a liquid across a pipe falls as a parabola from the centre to the wall.
- Doubling the output when the input doubles, in a square law.In \(y = kx^2\), multiplying \(x\) by \(n\) multiplies \(y\) by \(n^2\). Twice the speed gives four times the energy.
- Losing the sign in \(x_v = -b/2a\).Read \(a\) and \(b\) with their signs before you substitute. For \(20t - 5t^2\), \(a = -5\) and \(b = 20\).
- Keeping a negative root as a time of the motion.Check each root against the domain. A time before the throw or after the landing is not part of the flight.