Motion on a line: direction changes, distance and speed Problem set A Worked solutions

Each velocity is for an object moving on a straight line, in m s⁻¹, with time \(t \ge 0\) in seconds. Find the time or times when the object is at rest, and say whether it changes direction at each.

1. (a) \(v = t^{2} − 8t + 12\)

   • Polynomial solver: \(t^{2} − 8t + 12 = 0\), \(t = 2\) or \(t = 6\).
   • \(v(1) = 5\), \(v(4) = −4\), \(v(7) = 5\): the sign changes at each zero.
   • The object changes direction at \(t = 2\) and at \(t = 6\).

2. (b) \(v = t^{2} − 6t + 9\)

   • Polynomial solver: \(t^{2} − 6t + 9 = 0\), \(t = 3\) (a repeated solution).
   • \(v(2) = 1\) and \(v(4) = 1\): the same sign on both sides.
   • \(v\) touches zero but does not change sign, so the object stops for an instant and carries on the same way. No change of direction.

3. (c) \(v = 4\cos(0.5t)\) (first time only)

   • \(4\cos(0.5t) = 0\), \(0.5t = \dfrac{\pi}{2}\), \(t = 3.14\) s (radians).
   • \(v(1.57) = 2.83\), \(v(4.71) = −2.83\)
   • The sign changes, so the object changes direction.

A toy car moves along a straight track. Its velocity is \(v = t^{2} − 6t + 5\) m s⁻¹ at time \(t\) seconds.

1. (a) Find its displacement from the start in the first 6 seconds.

   • From the calculator: \(\int_0^{6} (t^{2} − 6t + 5)\,dt = −6\) m.

2. (b) Find the total distance it travels in the first 6 seconds.

   • It changes direction at \(t = 1\) and \(t = 5\) (v = 0).
   • From the calculator: \(\int_0^{6} \left|t^{2} − 6t + 5\right|\,dt = 15.3\) m.
   • Check: \(|2.33| + |−10.7| + |2.33| = 15.3\).

A model aircraft slides along a straight wire. Its velocity is \(v = 8\sin(0.5t) − 3\) m s⁻¹ for \(0 \le t \le 8\).

1. (a) Find the greatest velocity.

   • Acceleration \(a = 4\cos(0.5t)\), zero at \(t = 3.14\).
   • Compare the turning values with the ends: \(v(0) = −3\), \(v(3.14) = 5\), \(v(8) = −9.05\).
   • Greatest velocity 5 m s⁻¹.

2. (b) Find the least velocity.

   • Compare the same values: \(v(0) = −3\), \(v(3.14) = 5\), \(v(8) = −9.05\).
   • Least velocity −9.05 m s⁻¹.

3. (c) Find the greatest speed.

   • Speed is \(|v|\): the sizes are 3, 5, 9.05.
   • Greatest speed 9.05 m s⁻¹, at \(t = 8\).

A bead moves along a straight wire. Its displacement from the centre of the wire is \(s = 2t^{3} − 21t^{2} + 60t\) cm at time \(t\) seconds. Decide whether it is speeding up or slowing down:

1. (a) when \(t = 1\)

   • \(v = 6t^{2} − 42t + 60\), \(a = 12t − 42\).
   • \(v(1) = 24\), \(a(1) = −30\): opposite signs.
   • So the bead is slowing down.

2. (b) when \(t = 3\)

   • \(v = 6t^{2} − 42t + 60\), \(a = 12t − 42\).
   • \(v(3) = −12\), \(a(3) = −6\): the same sign.
   • So the bead is speeding up.

3. (c) when \(t = 4\)

   • \(v = 6t^{2} − 42t + 60\), \(a = 12t − 42\).
   • \(v(4) = −12\), \(a(4) = 6\): opposite signs.
   • So the bead is slowing down.

A trolley runs along a straight laboratory track. Its velocity is \(v = 2t^{2} − 14t + 24\) m s⁻¹ at time \(t\) seconds, for \(0 \le t \le 5\), where positive means towards the end stop.

1. (a) Find the times when it is at rest.

   • Polynomial solver: \(2t^{2} − 14t + 24 = 0\), \(t = 3\) or \(t = 4\).

2. (b) Find its displacement and the total distance it travels in the 5 seconds.

   • From the calculator: \(\int_0^{5} v\,dt = 28.3\) m.
   • From the calculator: \(\int_0^{5} |v|\,dt = 29\) m.

3. (c) Is it speeding up or slowing down when \(t = 3.2\)?

   • \(a = \dfrac{dv}{dt} = 4t − 14\)
   • \(v(3.2) = −0.32\), \(a(3.2) = −1.2\).
   • The signs are the same: speeding up.

A remote-controlled boat moves along a straight canal. The graph shows its velocity \(v\) in m s⁻¹ at time \(t\) seconds, where positive means towards the lock.

1. (a) Find the acceleration during the first 2 seconds.

   • The gradient of the first line: \(a = \dfrac{8 − 0}{2 − 0} = 4\) m s⁻².

2. (b) Find the time at which the boat changes direction.

   • The velocity changes sign where the line crosses the axis, which is at \(t = 9\).

3. (c) Find its displacement from the start at the end of the 13 seconds.

   • Area above the axis: 56 m. Area below the axis: 16 m.
   • Displacement \(56 − 16 = 40\) m.

4. (d) Find the total distance it travels in the 13 seconds.

   • Distance \(56 + 16 = 72\) m (both areas counted as positive).

A robot moves along a straight rail. The graph shows its displacement \(s\) metres from the charging point at time \(t\) seconds.

1. (a) Find the times when it is at rest.

   • It is at rest where the gradient is zero: the top and the bottom of the bends, at \(t = 2\) and \(t = 6\).

2. (b) Find its displacement from the start at \(t = 8\).

   • Read the height of the curve at \(t = 8\): 16 m.

3. (c) Find the total distance it travels in the 8 seconds.

   • Read the heights at the turning points: \(s(2) = 16\), \(s(6) = 0\).
   • Stages: 16 + 16 + 16 = 48 m.

4. (d) Find the time when it is back at its starting position.

   • The curve meets \(s = 0\) again at \(t = 6\).

A float on a river moves along a straight line. Its velocity is \(v = 6\sin(0.8t) − 2\) m s⁻¹ for \(0 \le t \le 8\). The calculator must be in radians.

1. (a) Find the first time it is at rest.

   • Solve \(6\sin(0.8t) − 2 = 0\) on the calculator: \(t = 0.425\) s (the first solution).

2. (b) Find the greatest speed in the interval.

   • Greatest and least velocity are at the turning points and the ends; compare the sizes of \(v(0) = −2\), \(v(1.96) = 4\), \(v(5.89) = −8\), \(v(8) = −1.30\).
   • Greatest speed 8 m s⁻¹.

3. (c) Find the total distance it travels in the 8 seconds.

   • It is at rest at \(t = 0.425\), \(t = 3.50\).
   • From the calculator: \(\int_0^{8} |6\sin(0.8t) − 2|\,dt = 31.9\) m.

A particle moves on a straight line, starting at the origin. Its velocity is \(v = 3t^{2} − 20t + 12\) m s⁻¹ at time \(t\) seconds.

1. (a) Find an expression for its displacement from the start at time \(t\).

   • \(s = \int v\,dt = t^{3} − 10t^{2} + 12t\) (the constant is zero because \(s = 0\) when \(t = 0\)).

2. (b) Find the first time after the start when it is back at its starting position.

   • Solve \(s = 0\): \(t^{3} − 10t^{2} + 12t = 0\), \(t = 0\), \(t = 1.39\), \(t = 8.61\).
   • The first time after the start is \(t = 1.39\) s.

3. (c) Find its velocity at that time. In which direction is it moving?

   • \(v(1.39) = −10.1\) m s⁻¹.
   • The velocity is negative, so it is moving in the negative direction.