The Workbook

1. Distance \(= 3 + 4 = 7\) km.

2. The two legs are perpendicular: \(|\vec s| = \sqrt{3^2 + 4^2} = 5\) km.

1. \(A \to B\) is \(8\) m and \(B \to C\) is \(5\) m: distance \(= 13\) m.

2. Displacement \(= x_C - x_A = 3 - 0 = +3\) m.

3. \(\dfrac{3}{13}\). The ratio can never exceed \(1\).

1. Distance \(= \pi r = \dfrac{22}{7} \times 7 = 22\) m. The body is at the opposite end of a diameter: displacement \(= 2r = 14\) m.

2. Distance \(= 2\pi r = 44\) m. The body is back at the start: displacement \(= 0\).

1. No. The straight line between two points is the shortest path, so the distance is at least the displacement.

2. Yes. On a round trip the body returns to the start: the displacement is zero but the distance is not.

Let each half be \(d\). Time \(= \dfrac{d}{v_1} + \dfrac{d}{v_2}\). Average speed \(= \dfrac{2d}{d/v_1 + d/v_2} = \dfrac{2v_1v_2}{v_1 + v_2}\).

\(= \dfrac{2 \times 40 \times 60}{100} = 48\) km/h. It is less than \(50\) because the car spends more time at the slower speed.

Let each half of the time be \(t\). Distance \(= 40t + 60t = 100t\). Average speed \(= \dfrac{100t}{2t} = 50\) km/h, the plain average.

In Question 5 the slow part took more time than the fast part; here both take the same time.

1. \(\dfrac{100}{50} = 2\) h and \(\dfrac{100}{25} = 4\) h.

2. \(\dfrac{200}{6} = \dfrac{100}{3} \approx 33\) km/h. Check with Question 5: \(\dfrac{2 \times 50 \times 25}{75} = \dfrac{100}{3}\).

Let each part be \(d\). Time \(= \dfrac{d}{20} + \dfrac{d}{30} + \dfrac{d}{60} = d\left(\dfrac{3 + 2 + 1}{60}\right) = \dfrac{d}{10}\).

Average speed \(= \dfrac{3d}{d/10} = 30\) km/h.

1. Distance \(= 14\) m in \(4\) s: \(3.5\) m/s.

2. Displacement \(= 10 - 4 = 6\) m east in \(4\) s: \(1.5\) m/s east.

1. \(0\) to \(2\) s: \(\dfrac{6 - 0}{2} = 3\) m/s. \(2\) to \(4\) s: \(\dfrac{2 - 6}{2} = -2\) m/s.

2. \(\dfrac{2 - 0}{4} = 0.5\) m/s.

3. Distance \(= 6 + 4 = 10\) m in \(4\) s: \(2.5\) m/s.

1. \(v = \dfrac{dx}{dt} = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)\). \(v = 0\) at \(t = 1\) s and \(t = 3\) s.

2. \(a = \dfrac{dv}{dt} = 6t - 12\). \(a = 0\) at \(t = 2\) s.

1. \(a = 4 - 2t = 2\) m/s\(^2\) at \(t = 1\) s.

2. \(v\) is greatest when \(a = 0\), at \(t = 2\) s (\(v = 4\) m/s).

3. \(\Delta x = \int_0^4 (4t - t^2)\,dt = \left[2t^2 - \dfrac{t^3}{3}\right]_0^4 = 32 - \dfrac{64}{3} = \dfrac{32}{3} \approx 10.7\) m.

\(v = \int_0^3 2t\,dt = [t^2]_0^3 = 9\) m/s. In general \(v = t^2\), so \(x = \int_0^3 t^2\,dt = \left[\dfrac{t^3}{3}\right]_0^3 = 9\) m.

\(\dfrac{dv}{dx} = 2\), so \(a = v\dfrac{dv}{dx} = 2x \times 2 = 4x\). At \(x = 3\) m: \(a = 12\) m/s\(^2\).

1. \(v\dfrac{dv}{dx} = -2v\) gives \(\dfrac{dv}{dx} = -2\). Integrate: \(v = -2x + C\), and \(v = u\) at \(x = 0\), so \(v = u - 2x\).

2. \(v = 0\) when \(x = \dfrac{u}{2} = 5\) m.

\(v\dfrac{dv}{dx} = 4x\), so \(v\,dv = 4x\,dx\). Integrate from the start: \(\dfrac{v^2}{2} = 2x^2\), hence \(v = 2x\). At \(x = 3\) m: \(v = 6\) m/s.

1. Intercept \(2\), slope \(\dfrac{10 - 2}{4} = 2\): \(v = 2 + 2x\).

2. \(\dfrac{dv}{dx} = 2\) s\(^{-1}\). At \(x = 1\) m, \(v = 4\) m/s, so \(a = 4 \times 2 = 8\) m/s\(^2\).

3. No. \(a = 2v\), and \(v\) grows with \(x\), so the acceleration grows too.

1. \(\int a\,dt\) is the change in velocity.

2. Area \(= \dfrac{1}{2} \times 3 \times 6 = 9\) m/s. \(v(3) = 2 + 9 = 11\) m/s.

3. \(a = 2t\). \(v = 2 + \int_0^t 2t\,dt = 2 + t^2\), which gives \(11\) m/s at \(t = 3\) s.

1. \(\bar v = \dfrac{x(3) - x(1)}{3 - 1} = \dfrac{18 - 2}{2} = 8\) m/s, the slope of the chord.

2. \(v = \dfrac{dx}{dt} = 4t = 8\) m/s at \(t = 2\) s, the slope of the tangent.

3. The acceleration is constant (\(4\) m/s\(^2\)), so \(v\) changes uniformly. The average over an interval then equals the instantaneous value at its midpoint, \(t = 2\) s.

1. Yes. A ball thrown up is at rest for an instant at the top, while its acceleration is still \(g\) downward.

2. Yes. In uniform circular motion the speed is constant but the direction of the velocity keeps changing.

Take right as positive: \(\Delta v = -5 - 5 = -10\) m/s. \(\bar a = \dfrac{-10}{2} = -5\) m/s\(^2\), that is, \(5\) m/s\(^2\) towards the left.

1. Distance \(= 150\) m in \(35\) s: \(\dfrac{150}{35} = \dfrac{30}{7} \approx 4.3\) m/s.

2. Displacement \(= 50\) m in \(35\) s: \(\dfrac{50}{35} = \dfrac{10}{7} \approx 1.4\) m/s, in the direction of the run.

1. \(\dfrac{30 - 10}{5} = 4\) m/s\(^2\).

2. \(\dfrac{0 - 20}{10} = -2\) m/s\(^2\). The minus sign means the acceleration opposes the motion.

1. At \(t = 0\), \(x = 8\) m. \(v = 4 - 4t\), so \(u = 4\) m/s. \(a = -4\) m/s\(^2\).

2. \(v = 0\) at \(t = 1\) s. Then \(x = 8 + 4 - 2 = 10\) m.

1. \(v = u + at = 5 + 2 \times 4 = 13\) m/s.

2. \(s = ut + \dfrac{1}{2}at^2 = 20 + \dfrac{1}{2} \times 2 \times 16 = 36\) m.

1. \(0 = 20 - 4t\) gives \(t = 5\) s.

2. \(0 = 20^2 - 2 \times 4 \times s\) gives \(s = \dfrac{400}{8} = 50\) m.

\(v^2 = u^2 + 2as\): \(900 = 100 + 80a\), so \(a = 10\) m/s\(^2\). Then \(t = \dfrac{v - u}{a} = \dfrac{20}{10} = 2\) s.

\(s = \dfrac{1}{2}at^2\): \(100 = 50a\), so \(a = 2\) m/s\(^2\). \(v = at = 20\) m/s.

1. The slope of the \(v\)-\(t\) line is the acceleration: \(a = \dfrac{v - u}{t}\), so \(v = u + at\).

2. The displacement is the area under the graph: rectangle \(ut\) plus triangle \(\dfrac{1}{2}(v - u)t = \dfrac{1}{2}(at)t\). So \(s = ut + \dfrac{1}{2}at^2\).

3. \(t = \dfrac{v - u}{a}\). Then \(s = \dfrac{u(v - u)}{a} + \dfrac{(v - u)^2}{2a} = \dfrac{(v - u)(2u + v - u)}{2a} = \dfrac{v^2 - u^2}{2a}\), so \(v^2 = u^2 + 2as\).

1. \(0\) to \(4\) s: \(\dfrac{20}{4} = 5\) m/s\(^2\). \(4\) to \(10\) s: \(0\). \(10\) to \(12\) s: \(\dfrac{-20}{2} = -10\) m/s\(^2\).

2. Areas: \(\dfrac{1}{2} \times 4 \times 20 + 6 \times 20 + \dfrac{1}{2} \times 2 \times 20 = 40 + 120 + 20 = 180\) m.

3. \(\dfrac{180}{12} = 15\) m/s.

\(s = ut + \dfrac{1}{2}at^2\). At \(t = 1\): \(u + \dfrac{a}{2} = 3\). At \(t = 2\): \(2u + 2a = 10\), that is, \(u + a = 5\).

Subtract: \(\dfrac{a}{2} = 2\), so \(a = 4\) m/s\(^2\) and \(u = 1\) m/s.

\(n = 2\): \(u + \dfrac{3a}{2} = 20\). \(n = 4\): \(u + \dfrac{7a}{2} = 30\). Subtract: \(2a = 10\), so \(a = 5\) m/s\(^2\) and \(u = 20 - 7.5 = 12.5\) m/s.

Distance in \(n\) seconds: \(s(n) = un + \dfrac{1}{2}an^2\). In \((n - 1)\) seconds: \(s(n - 1) = u(n - 1) + \dfrac{1}{2}a(n - 1)^2\).

\(s_n = s(n) - s(n - 1) = u + \dfrac{a}{2}\left[n^2 - (n - 1)^2\right] = u + \dfrac{a}{2}(2n - 1)\).

1. \(s_5 = 0 + \dfrac{2}{2}(2 \times 5 - 1) = 9\) m.

2. \(s = \dfrac{1}{2} \times 2 \times 25 = 25\) m. Check: \(1 + 3 + 5 + 7 + 9 = 25\).

With \(u = 0\), \(s_n = \dfrac{a}{2}(2n - 1)\), so \(s_1 : s_2 : s_3 = 1 : 3 : 5\). The distance in the \(3\)rd second is \(5 \times 3 = 15\) m.

1. \(v = at = 2 \times 10 = 20\) m/s.

2. Braking: \(0 = 20 - 5t\), so \(t = 4\) s. Total time \(= 14\) s.

3. Area of the triangle \(= \dfrac{1}{2} \times 14 \times 20 = 140\) m. Check: \(\dfrac{1}{2} \times 2 \times 100 + \dfrac{20^2}{2 \times 5} = 100 + 40 = 140\) m.

Stage 1: \(s = \dfrac{1}{2} \times 1 \times 16 = 8\) m, reaching \(v = 4\) m/s. Stage 2: \(s = 4 \times 6 = 24\) m. Stage 3: \(s = \dfrac{v^2}{2b} = \dfrac{16}{4} = 4\) m. Total \(= 8 + 24 + 4 = 36\) m.

1. The speed rises from \(0\) to \(v\) in \(t_1\) at rate \(a\): \(v = at_1\). It falls from \(v\) to \(0\) in \(t_2\) at rate \(b\): \(v = bt_2\). The \(v\)-\(t\) graph is a triangle of base \(t_1 + t_2\) and height \(v\), so the distance is \(\dfrac{1}{2}v(t_1 + t_2)\).

2. \(t_1 = 2t_2\) (from \(at_1 = bt_2\)) and \(t_1 + t_2 = 6\): \(t_1 = 4\) s, \(t_2 = 2\) s. \(v = 1 \times 4 = 4\) m/s. Distance \(= \dfrac{1}{2} \times 4 \times 6 = 12\) m.

Stage 1: \(s = 10 \times 5 + \dfrac{1}{2} \times 2 \times 25 = 75\) m, and \(v = 10 + 10 = 20\) m/s. Stage 2: \(20 \times 5 = 100\) m. Total \(= 175\) m.

1. \(0 = 20^2 - 2 \times 5 \times s\), so \(s = \dfrac{400}{10} = 40\) m.

2. \(s = \dfrac{1600}{10} = 160\) m, four times as much. Doubling the speed multiplies the stopping distance by four.

3. With \(v = 0\): \(0 = u^2 - 2as\), so \(s = \dfrac{u^2}{2a}\).

1. The car moves at a steady \(20\) m/s for \(0.5\) s: \(20 \times 0.5 = 10\) m.

2. \(\dfrac{u^2}{2a} = \dfrac{400}{10} = 40\) m.

3. \(10 + 40 = 50\) m, the whole area under the graph.

At \(20\) m/s the stopping distance is \(50\) m, less than \(60\) m: the car stops in time.

At \(25\) m/s: reaction \(25 \times 0.5 = 12.5\) m, braking \(\dfrac{625}{10} = 62.5\) m, total \(75\) m. That is more than \(60\) m, so the car hits the obstacle.

\(s = \dfrac{u^2}{2a}\): \(u^2 = 2 \times 5 \times 90 = 900\), so \(u = 30\) m/s.

Let \(AB = s\). Over the whole length \(v^2 = u^2 + 2as\). Over half the length \(v_m^2 = u^2 + as = u^2 + \dfrac{v^2 - u^2}{2} = \dfrac{u^2 + v^2}{2}\).

\(v_m = \sqrt{\dfrac{1 + 49}{2}} = \sqrt{25} = 5\) m/s. Note that this is more than the plain average \(4\) m/s.

\(\bar v = \dfrac{4 + 16}{2} = 10\) m/s. \(s = \bar v t = 10 \times 6 = 60\) m.

First \(2\) s: \(2u + 2a = 10\), so \(u + a = 5\). First \(4\) s: \(4u + 8a = 30\), so \(u + 2a = 7.5\). Subtract: \(a = 2.5\) m/s\(^2\) and \(u = 2.5\) m/s.

\(v^2 = u^2 + 2as\): \(400 = 100 + 150a\), so \(a = 2\) m/s\(^2\). \(t = \dfrac{20 - 10}{2} = 5\) s.

\(0 = 200^2 - 2a \times 0.10\), so \(a = \dfrac{40000}{0.2} = 2 \times 10^{5}\) m/s\(^2\). \(t = \dfrac{u}{a} = \dfrac{200}{2 \times 10^{5}} = 10^{-3}\) s.

1. \(v = gt = 10 \times 2 = 20\) m/s.

2. \(s = \dfrac{1}{2}gt^2 = 5 \times 4 = 20\) m.

3. \(s = 5 \times 9 = 45\) m.

\(80 = 5t^2\) gives \(t = 4\) s. \(v = gt = 40\) m/s. Check: \(v^2 = 2gh = 1600\).

\(s_n = \dfrac{g}{2}(2n - 1) = 5(2n - 1)\): \(5\), \(15\), \(25\) and \(35\) m. Each second adds \(10\) m more than the one before.

\(h = \dfrac{1}{2}gt^2 = 5 \times 9 = 45\) m. \(v = gt = 30\) m/s.

1. First second: \(5\) m. Second: \(20 - 5 = 15\) m. Third: \(45 - 20 = 25\) m.

2. \(5 : 15 : 25 = 1 : 3 : 5\), the odd numbers, as \(s_n = 5(2n - 1)\) predicts.

3. \(v = gt\): \(10\) m/s at \(1\) s, \(20\) m/s at \(2\) s, \(30\) m/s at \(3\) s.

1. \(0 = 20 - 10t\) gives \(t = 2\) s.

2. \(H = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20\) m.

3. The way down takes as long as the way up: \(4\) s in all. The ball returns at \(20\) m/s, the speed of the throw.

\(h = 30t - 5t^2\). At \(t = 1\) s: \(30 - 5 = 25\) m. At \(t = 5\) s: \(150 - 125 = 25\) m.

The ball reaches the top at \(t = 3\) s. The instants \(1\) s and \(5\) s are \(2\) s before and \(2\) s after the top. So the ball is at the same height, once going up and once coming down.

\(60 = 40t - 5t^2\), so \(t^2 - 8t + 12 = 0\) and \((t - 2)(t - 6) = 0\): \(t = 2\) s and \(t = 6\) s.

The ball passes \(60\) m on the way up at \(2\) s and again on the way down at \(6\) s. The top is at \(t = 4\) s, midway.

1. The line crosses the axis at \(t = 2\) s: there \(v = 0\), the top of the flight.

2. Slope \(= \dfrac{-20 - 20}{4} = -10\) m/s\(^2\): the acceleration, \(-g\), the same throughout.

3. \(0\) to \(2\) s: \(\dfrac{1}{2} \times 2 \times 20 = +20\) m, the rise to the top. \(2\) to \(4\) s: \(-20\) m, the fall back. The net displacement is zero.

With up positive the acceleration is \(-g\): \(v = u - gt\) and \(h = ut - \dfrac{1}{2}gt^2\).

At the top the velocity is zero for an instant, but the acceleration is still \(-g\): the ball does not stay there.

Take the roof as the origin and up as positive. The ground is at \(h = -25\) m: \(-25 = 20t - 5t^2\), so \(t^2 - 4t - 5 = 0\) and \((t - 5)(t + 1) = 0\): \(t = 5\) s.

\(v = 20 - 50 = -30\) m/s: the ball arrives at \(30\) m/s, moving down.

Take down as positive: \(75 = 10t + 5t^2\), so \(t^2 + 2t - 15 = 0\) and \((t + 5)(t - 3) = 0\): \(t = 3\) s. \(v = 10 + 30 = 40\) m/s.

1. \(A\): \(h_A = 60 - 5t^2\). \(B\): \(h_B = 20t - 5t^2\).

2. \(h_A = h_B\) gives \(60 = 20t\), so \(t = 3\) s. Then \(h = 60 - 45 = 15\) m above the ground.

3. \(h - \dfrac{1}{2}gt^2 = ut - \dfrac{1}{2}gt^2\) gives \(t = \dfrac{h}{u}\). The \(\dfrac{1}{2}gt^2\) terms cancel: gravity affects both balls equally, so they approach each other at the steady relative speed \(u\).

\(15 = ut - 5t^2\), so \(5t^2 - ut + 15 = 0\). Its roots \(t_1\), \(t_2\) satisfy \(t_1 + t_2 = \dfrac{u}{5}\) and \(t_1t_2 = 3\).

\((t_2 - t_1)^2 = (t_1 + t_2)^2 - 4t_1t_2\): \(4 = \dfrac{u^2}{25} - 12\), so \(u^2 = 400\) and \(u = 20\) m/s. Check: the roots are \(1\) s and \(3\) s.

1. \(10\) m/s upward. The ball shares the balloon's velocity at the moment of release; only after that does gravity act on it alone.

2. Up positive, ground at \(-75\) m: \(-75 = 10t - 5t^2\), so \(t^2 - 2t - 15 = 0\) and \((t - 5)(t + 3) = 0\): \(t = 5\) s.

3. \(v = 10 - 50 = -40\) m/s: \(40\) m/s downward.

\(5(2n - 1) = 45\) gives \(2n - 1 = 9\), so \(n = 5\): the fifth second.

The last second is the \(n\)th second with \(5(2n - 1) = 45\), so \(n = 5\): the fall lasts \(5\) s. Height \(= \dfrac{1}{2} \times 10 \times 25 = 125\) m.

At the top \(v = 0\): \(0 = u - gT\), so \(u = gT\). Then \(H = uT - \dfrac{1}{2}gT^2 = gT^2 - \dfrac{1}{2}gT^2 = \dfrac{1}{2}gT^2\).

\(T = 3\) s: \(u = 30\) m/s and \(H = 5 \times 9 = 45\) m.

Time to the top: \(0 = u - gt\), \(t = \dfrac{u}{g}\); the whole flight takes twice that, \(\dfrac{2u}{g}\). Height: \(0 = u^2 - 2gH\), so \(H = \dfrac{u^2}{2g}\).

Doubling \(u\) doubles the time of flight and multiplies the height by four.

1. The top of the curve is at \(20\) m, at \(t = 2\) s. The curve returns to zero at \(t = 4\) s: the time of flight.

2. \(1\) s and \(3\) s. Check: \(20 - 5 = 15\) and \(60 - 45 = 15\).

3. The slope of \(h\) against \(t\) is the velocity. It is zero at the top, \(t = 2\) s.

\(s = 5 \times 4 = 20\) m and \(v = 10 \times 2 = 20\) m/s.

1. The first drop falls \(5\) m in \(t\) with \(5 = 5t^2\), so \(t = 1\) s. Two intervals fit in this time: the interval is \(0.5\) s.

2. The second drop has fallen for \(0.5\) s: \(5 \times 0.25 = 1.25\) m. It is \(5 - 1.25 = 3.75\) m above the floor.

Moon: \(H = \dfrac{64}{2 \times 1.6} = 20\) m and \(T = \dfrac{2 \times 8}{1.6} = 10\) s.

Earth: \(H = \dfrac{64}{20} = 3.2\) m and \(T = \dfrac{16}{10} = 1.6\) s. On the Moon the ball goes about six times higher and stays up six times longer.

\(v = \sqrt{2gh}\), so four times the height gives \(\sqrt{4} = 2\) times the speed: \(2v\). \(t = \sqrt{\dfrac{2h}{g}}\), so the time also doubles: \(2t\).