Motion on a line: direction changes, distance and speed Problem set B 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) \(v = t^{2} − 10t + 21\)

   • Polynomial solver: \(t^{2} − 10t + 21 = 0\), \(t = 3\) or \(t = 7\).
   • \(v(2) = 5\), \(v(5) = −4\), \(v(8) = 5\): the sign changes at each zero.
   • The object changes direction at \(t = 3\) and at \(t = 7\).

2. (b) \(v = t^{2} − 10t + 25\)

   • Polynomial solver: \(t^{2} − 10t + 25 = 0\), \(t = 5\) (a repeated solution).
   • \(v(4) = 1\) and \(v(6) = 1\): 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) \(v = 6\cos(2t)\) (first time only)

   • \(6\cos(2t) = 0\), \(2t = \dfrac{\pi}{2}\), \(t = 0.785\) s (radians).
   • \(v(0.393) = 4.24\), \(v(1.18) = −4.24\)
   • The sign changes, so the object changes direction.

A tram moves along a straight street. Its velocity is \(v = t^{2} − 9t + 14\) m s⁻¹ at time \(t\) seconds.

1. (a) Find its displacement from the start in the first 8 seconds.

   • From the calculator: \(\int_0^{8} (t^{2} − 9t + 14)\,dt = −5.33\) m.

2. (b) Find the total distance it travels in the first 8 seconds.

   • It changes direction at \(t = 2\) and \(t = 7\) (v = 0).
   • From the calculator: \(\int_0^{8} \left|t^{2} − 9t + 14\right|\,dt = 36.3\) m.
   • Check: \(|12.7| + |−20.8| + |2.83| = 36.3\).

A pendulum bob moves along a straight line. Its velocity is \(v = 12\cos(0.4t) − 5\) m s⁻¹ for \(0 \le t \le 10\).

1. (a) Find the greatest velocity.

   • Acceleration \(a = −4.8\sin(0.4t)\), zero at \(t = 0\) and \(t = 7.85\).
   • Compare the turning values with the ends: \(v(0) = 7\), \(v(7.85) = −17\), \(v(10) = −12.8\).
   • Greatest velocity 7 m s⁻¹.

2. (b) Find the least velocity.

   • Compare the same values: \(v(0) = 7\), \(v(7.85) = −17\), \(v(10) = −12.8\).
   • Least velocity −17 m s⁻¹.

3. (c) Find the greatest speed.

   • Speed is \(|v|\): the sizes are 7, 17, 12.8.
   • Greatest speed 17 m s⁻¹, at \(t = 7.85\).

A piston moves along a straight cylinder. Its displacement from the base of the cylinder is \(s = t^{3} − 15t^{2} + 63t\) cm at time \(t\) seconds. Decide whether it is speeding up or slowing down:

1. (a) when \(t = 2\)

   • \(v = 3t^{2} − 30t + 63\), \(a = 6t − 30\).
   • \(v(2) = 15\), \(a(2) = −18\): opposite signs.
   • So the piston is slowing down.

2. (b) when \(t = 4\)

   • \(v = 3t^{2} − 30t + 63\), \(a = 6t − 30\).
   • \(v(4) = −9\), \(a(4) = −6\): the same sign.
   • So the piston is speeding up.

3. (c) when \(t = 6\)

   • \(v = 3t^{2} − 30t + 63\), \(a = 6t − 30\).
   • \(v(6) = −9\), \(a(6) = 6\): opposite signs.
   • So the piston is slowing down.

A lift moves up and down a straight shaft. Its velocity is \(v = 3t^{2} − 21t + 30\) m s⁻¹ at time \(t\) seconds, for \(0 \le t \le 6\), where positive means upwards.

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

   • Polynomial solver: \(3t^{2} − 21t + 30 = 0\), \(t = 2\) or \(t = 5\).

2. (b) Find its displacement and the total distance it travels in the 6 seconds.

   • From the calculator: \(\int_0^{6} v\,dt = 18\) m.
   • From the calculator: \(\int_0^{6} |v|\,dt = 45\) m.

3. (c) Is it speeding up or slowing down when \(t = 4.5\)?

   • \(a = \dfrac{dv}{dt} = 6t − 21\)
   • \(v(4.5) = −3.75\), \(a(4.5) = 6\).
   • The signs are opposite: slowing down.

A test car drives along a straight track. The graph shows its velocity \(v\) in m s⁻¹ at time \(t\) seconds, where positive means forwards.

1. (a) Find the acceleration during the first 4 seconds.

   • The gradient of the first line: \(a = \dfrac{12 − 0}{4 − 0} = 3\) m s⁻².

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

   • The velocity changes sign where the line crosses the axis, which is at \(t = 11\).

3. (c) Find its displacement from the start at the end of the 14 seconds.

   • Area above the axis: 90 m. Area below the axis: 6 m.
   • Displacement \(90 − 6 = 84\) m.

4. (d) Find the total distance it travels in the 14 seconds.

   • Distance \(90 + 6 = 96\) m (both areas counted as positive).

A toy train runs along a straight track. The graph shows its displacement \(s\) metres from the station 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 \(t = 3\) and \(t = 9\).

2. (b) Find its displacement from the start at \(t = 12\).

   • Read the height of the curve at \(t = 12\): 36 m.

3. (c) Find the total distance it travels in the 12 seconds.

   • Read the heights at the turning points: \(s(3) = 36\), \(s(9) = 0\).
   • Stages: 36 + 36 + 36 = 108 m.

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

   • The curve meets \(s = 0\) again at \(t = 9\).

A cable car moves along a straight line. Its velocity is \(v = 9\sin(0.5t) − 3\) m s⁻¹ for \(0 \le t \le 10\). The calculator must be in radians.

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

   • Solve \(9\sin(0.5t) − 3 = 0\) on the calculator: \(t = 0.680\) 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 \(v(0) = −3\), \(v(3.14) = 6\), \(v(9.42) = −12\), \(v(10) = −11.6\).
   • Greatest speed 12 m s⁻¹.

3. (c) Find the total distance it travels in the 10 seconds.

   • It is at rest at \(t = 0.680\), \(t = 5.60\).
   • From the calculator: \(\int_0^{10} |9\sin(0.5t) − 3|\,dt = 55.4\) m.

A block slides on a straight rail, starting at the origin. Its velocity is \(v = 3t^{2} − 26t + 16\) m s⁻¹ at time \(t\) seconds.

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

   • \(s = \int v\,dt = t^{3} − 13t^{2} + 16t\) (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 \(s = 0\): \(t^{3} − 13t^{2} + 16t = 0\), \(t = 0\), \(t = 1.38\), \(t = 11.6\).
   • The first time after the start is \(t = 1.38\) s.

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

   • \(v(1.38) = −14.1\) m s⁻¹.
   • The velocity is negative, so it is moving in the negative direction.