Multi-stage motion with velocity-time graph given

\includegraphics{figure_6} The diagram shows the \((t, v)\) graph for the motion of a hoist used to deliver materials to different levels at a building site. The hoist moves vertically. The graph consists of straight line segments. In the first stage the hoist travels upwards from ground level for 25 s, coming to rest 8 m above ground level.

1. Find the greatest speed reached by the hoist during this stage. [2]

The second stage consists of a 40 s wait at the level reached during the first stage. In the third stage the hoist continues upwards until it comes to rest 40 m above ground level, arriving 135 s after leaving ground level. The hoist accelerates at \(0.02 \text{ m s}^{-2}\) for the first 40 s of the third stage, reaching a speed of \(V \text{ m s}^{-1}\). Find

1. the value of \(V\), [3]
2. the length of time during the third stage for which the hoist is moving at constant speed, [4]
3. the deceleration of the hoist in the final part of the third stage. [3]

A particle travels in a straight line during the time interval \(0 \leqslant t \leqslant 12\), where \(t\) is the time in seconds. Fig. 1 is the velocity-time graph for the motion. \includegraphics{figure_3}

1. Calculate the acceleration of the particle in the interval \(0 < t < 6\). [2]
2. Calculate the distance travelled by the particle from \(t = 0\) to \(t = 4\). [2]
3. When \(t = 0\) the particle is at A. Calculate how close the particle gets to A during the interval \(4 \leqslant t \leqslant 12\). [2]

In this question take \(g\) as \(10\text{ m s}^{-2}\). A small ball is released from rest. It falls for 2 seconds and is then brought to rest over the next 5 seconds. This motion is modelled in the speed-time graph Fig. 6. \includegraphics{figure_4} For this model,

1. calculate the distance fallen from \(t = 0\) to \(t = 7\), [3]
2. find the acceleration of the ball from \(t = 2\) to \(t = 6\), specifying the direction, [3]
3. obtain an expression in terms of \(t\) for the downward speed of the ball from \(t = 2\) to \(t = 6\), [3]
4. state the assumption that has been made about the resistance to motion from \(t = 0\) to \(t = 2\). [1]

The part of the motion from \(t = 2\) to \(t = 7\) is now modelled by \(v = -\frac{3}{2}t^2 + \frac{19}{2}t + 7\).

1. Verify that \(v\) agrees with the values given in Fig. 6 at \(t = 2\), \(t = 6\) and \(t = 7\). [2]
2. Calculate the distance fallen from \(t = 2\) to \(t = 7\) according to this model. [7]

The graph shows how the speed of a cyclist varies during a timed section of length 120 metres along a straight track. \includegraphics{figure_15}

1. Find the acceleration of the cyclist during the first 10 seconds. [1 mark]
2. After the first 15 seconds, the cyclist travels at a constant speed of 5 m s⁻¹ for a further \(T\) seconds to complete the 120-metre section. Calculate the value of \(T\). [4 marks]