Particle motion - velocity/time (dv/dt = f(v,t))

6 A cyclist and her bicycle have a total mass of 60 kg . The cyclist rides in a horizontal straight line, and exerts a constant force in the direction of motion of 150 N . The motion is opposed by a resistance of magnitude \(12 v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the cyclist's speed at time \(t \mathrm {~s}\) after passing through a fixed point \(A\).

1. Show that \(5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = 12.5 - v\).
2. Given that the cyclist passes through \(A\) with speed \(11.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), solve this differential equation to show that \(v = 12.5 - \mathrm { e } ^ { - 0.2 t }\).
3. Express the displacement of the cyclist from \(A\) in terms of \(t\).

6 A cyclist and his machine have a total mass of 80 kg . The cyclist starts from rest and rides from the bottom to the top of a straight slope inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = 0.1\). The cyclist exerts a constant force of magnitude 120 N . There is a resisting force of magnitude \(8 v \mathrm {~N}\) acting on the cyclist, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the cyclist's speed at time \(t \mathrm {~s}\) after the start.

1. Show that \(\left( \frac { 1 } { 5 - v } \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 1 } { 10 }\).
2. Solve this differential equation and hence show that \(v = 5 \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 10 } t } \right)\).
3. Given that the cyclist takes 20 s to reach the top of the slope, find the length of the slope.

6 A cyclist and his bicycle have a total mass of 81 kg . The cyclist starts from rest and rides in a straight line. The cyclist exerts a constant force of 135 N and the motion is opposed by a resistance of magnitude \(9 v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the cyclist's speed at time \(t \mathrm {~s}\) after starting.

1. Show that \(\frac { 9 } { 15 - v } \frac { \mathrm {~d} v } { \mathrm {~d} t } = 1\).
2. Solve this differential equation to show that \(v = 15 \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 9 } t } \right)\).
3. Find the distance travelled by the cyclist in the first 9 s of the motion.

8 Vicky has mass 65 kg and is skydiving. She steps out of a helicopter and falls vertically. She then waits a short period of time before opening her parachute. The parachute opens at time \(t = 0\) when her speed is \(19.6 \mathrm {~ms} ^ { - 1 }\), and she then experiences an air resistance force of magnitude \(260 v\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is her speed at time \(t\) seconds.

1. When \(t > 0\) :
   1. show that the resultant downward force acting on Vicky is 65(9.8-4v) newtons
   2. show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 4 ( v - 2.45 )\).
2. By showing that \(\int \frac { 1 } { v - 2.45 } \mathrm {~d} v = - \int 4 \mathrm {~d} t\), find \(v\) in terms of \(t\).
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7 A stone, of mass 5 kg , is projected vertically downwards, in a viscous liquid, with an initial speed of \(7 \mathrm {~ms} ^ { - 1 }\). At time \(t\) seconds after it is projected, the stone has speed \(v \mathrm {~ms} ^ { - 1 }\) and it experiences a resistance force of magnitude \(9.8 v\) newtons.

1. When \(t \geqslant 0\), show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 1.96 ( v - 5 )$$ (2 marks)
2. Find \(v\) in terms of \(t\).

A train is moving along a straight horizontal track. At time t seconds, its velocity is \(v \mathrm {~ms} ^ { - 1 }\), its acceleration is \(a \mathrm {~ms} ^ { - 2 }\), and \(a\) is inversely proportional to V . At time \(\mathrm { t } = 1\), \(v = 5\) and \(a = 1 \cdot 8\). a) i) Write down a differential equation satisfied by V.
ii) Show that \(v ^ { 2 } = 18 t + 7\).
b) Find the time at which the magnitude of the velocity is equal to the magnitude of the acceleration. \section*{END OF PAPER}

A car \(C\) is moving horizontally in a straight line with velocity \(v \text{ms}^{-1}\) at time \(t\) seconds, where \(v > 0\) and \(t \geq 0\). The acceleration, \(a \text{ms}^{-2}\), of \(C\) is modelled by the equation $$a = v\left(\frac{8t}{7 + 4t^2} - \frac{1}{2}\right).$$

1. In this question you must show detailed reasoning. Find the times when the acceleration of \(C\) is zero. [3] At \(t = 0\) the velocity of \(C\) is \(17.5 \text{ms}^{-1}\) and at \(t = T\) the velocity of \(C\) is \(5 \text{ms}^{-1}\).
2. By setting up and solving a differential equation, show that \(T\) satisfies the equation $$T = 2 \ln\left(\frac{7 + 4T^2}{2}\right).$$ [6]
3. Use an iterative formula, based on the equation in part (b), to find the value of \(T\), giving your answer correct to 4 significant figures. Use an initial value of 11.25 and show the result of each step of the iteration process. [2]
4. The diagram below shows the velocity-time graph for the motion of \(C\). \includegraphics{figure_7d} Find the time taken for \(C\) to decelerate from travelling at its maximum speed until it is travelling at \(5 \text{ms}^{-1}\). [1]

A particle moves so that its acceleration, \(a\text{ ms}^{-2}\), at time \(t\) seconds may be modelled in terms of its velocity, \(v\text{ ms}^{-1}\), as $$a = -0.1v^2$$ The initial velocity of the particle is \(4\text{ ms}^{-1}\)

1. By first forming a suitable differential equation, show that $$v = \frac{20}{5 + 2t}$$ [6 marks]
2. Find the acceleration of the particle when \(t = 5.5\) [2 marks]

An object of mass \(0 \cdot 5\) kg is thrown vertically upwards with initial speed \(24\) ms\(^{-1}\). The velocity of the object at time \(t\) seconds is \(v\) ms\(^{-1}\). During the upward motion, the object experiences a resistance to motion \(RN\), where \(R\) is proportional to \(v\). When the velocity of the object is \(0 \cdot 2\) ms\(^{-1}\) the resistance to motion is \(0 \cdot 08\) N.

1. Show that the upward motion of the object satisfies the differential equation $$\frac{\mathrm{d}v}{\mathrm{d}t} = -9 \cdot 8 - 0 \cdot 8\,v.$$ [3]
2. Find an expression for \(v\) at time \(t\). [6]
3. Determine the value of \(t\) when the object is at the highest point of the motion. [2]