Vertical motion: velocity-time graph

3 \includegraphics[max width=\textwidth, alt={}, center]{d3bb6702-231d-42a0-830e-9f844dca78d7-2_748_1410_979_370} The velocity-time graph shown models the motion of a parachutist falling vertically. There are four stages in the motion:

• falling freely with the parachute closed,
• decelerating at a constant rate with the parachute open,
• falling with constant speed with the parachute open,
• coming to rest instantaneously on hitting the ground.
  1. Show that the total distance fallen is 1048 m .

The weight of the parachutist is 850 N .

Find the upward force on the parachutist due to the parachute, during the second stage.

1 \includegraphics[max width=\textwidth, alt={}, center]{5125fab5-0be5-4904-afdf-93e91b16e773-2_608_831_258_657} Two particles \(P\) and \(Q\) move vertically under gravity. The graphs show the upward velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) of the particles at time \(t \mathrm {~s}\), for \(0 \leqslant t \leqslant 4 . P\) starts with velocity \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(Q\) starts from rest.

1. Find the value of \(V\). Given that \(Q\) reaches the horizontal ground when \(t = 4\), find
2. the speed with which \(Q\) reaches the ground,
3. the height of \(Q\) above the ground when \(t = 0\).

7. A helicopter is hovering at rest above horizontal ground at the point \(H\). A parachutist steps out of the helicopter and immediately falls vertically and freely under gravity from rest for 2.5 s . His parachute then opens and causes him to immediately decelerate at a constant rate of \(3.9 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) for \(T\) seconds ( \(T < 6\) ), until his speed is reduced to \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). He then moves with this constant speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) until he hits the ground. While he is decelerating, he falls a distance of 73.75 m . The total time between the instant when he leaves \(H\) and the instant when he hits the ground is 20 s . The parachutist is modelled as a particle.

1. Find the speed of the parachutist at the instant when his parachute opens.
2. Sketch a speed-time graph for the motion of the parachutist from the instant when he leaves \(H\) to the instant when he hits the ground.
3. Find the value of \(T\).
4. Find, to the nearest metre, the height of the point \(H\) above the ground.
   7. A helicopter is hovering at rest above horizontal ground at the point \(H\). A parachutist steps

6. A parachutist drops from a helicopter \(H\) and falls vertically from rest towards the ground. Her parachute opens 2 s after she leaves \(H\) and her speed then reduces to \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). For the first 2 s her motion is modelled as that of a particle falling freely under gravity. For the next 5 s the model is motion with constant deceleration, so that her speed is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at the end of this period. For the rest of the time before she reaches the ground, the model is motion with constant speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Sketch a speed-time graph to illustrate her motion from \(H\) to the ground.
2. Find her speed when the parachute opens. A safety rule states that the helicopter must be high enough to allow the parachute to open and for the speed of a parachutist to reduce to \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) before reaching the ground. Using the assumptions made in the above model,
3. find the minimum height of \(H\) for which the woman can make a drop without breaking this safety rule. Given that \(H\) is 125 m above the ground when the woman starts her drop,
4. find the total time taken for her to reach the ground.
5. State one way in which the model could be refined to make it more realistic.
   (1 mark)

1 An object C is moving along a vertical straight line. Fig. 1 shows the velocity-time graph for part of its motion. Initially C is moving upwards at \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and after 10 s it is moving downwards at \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \begin{figure}[h] \end{figure} C then moves as follows.

• In the interval \(10 \leqslant t \leqslant 15\), the velocity of C is constant at \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards.
• In the interval \(15 \leqslant t \leqslant 20\), the velocity of C increases uniformly so that C has zero velocity at \(t = 20\).
  1. Complete the velocity-time graph for the motion of C in the time interval \(0 \leqslant t \leqslant 20\).
  2. Calculate the acceleration of C in the time interval \(0 < t < 10\).
  3. Calculate the displacement of C from \(t = 0\) to \(t = 20\).

6. A student attempts to sketch the acceleration-time graph of a parachutist who jumps from a plane at a height of 2200 m above the ground. The student assumes that the parachutist falls freely from rest under gravity until she is 240 m from the ground at which point she opens her parachute. The student makes the assumption that, at this point, the velocity of the parachutist is immediately reduced to a value which remains constant until she reaches the ground 140 seconds after she left the plane. \begin{figure}[h] \end{figure} The student decides to ignore air resistance and his sketch is shown in Figure 3. The value \(t _ { 1 }\) is used by the student to denote the time at which the parachute is opened. Using the model proposed by the student, calculate

1. the speed of the parachutist immediately before she opens her parachute,
2. the value of \(t _ { 1 }\),
3. the speed of the parachutist after the parachute is opened.
4. Comment on two features of the student's model which are unrealistic and say what effect taking account of these would have had on the values which you calculated in parts (a) and (b).
   (4 marks)

The graph below shows the velocity of an object moving in a straight line over a 20 second journey. \includegraphics{figure_4}

1. Find the maximum magnitude of the acceleration of the object. [1 mark]
2. The object is at its starting position at times 0, \(t_1\) and \(t_2\) seconds. Find \(t_1\) and \(t_2\) [4 marks]

The diagram below shows a velocity-time graph for a particle moving with velocity \(v \text{ m s}^{-1}\) at time \(t\) seconds. \includegraphics{figure_10} Which statement is correct? Tick (\(\checkmark\)) one box. [1 mark] The particle was stationary for \(9 \leq t \leq 12\) The particle was decelerating for \(12 \leq t \leq 20\) The particle had a displacement of zero when \(t = 6\) The particle's speed when \(t = 4\) was \(-12 \text{ m s}^{-1}\)