6.06a Variable force: dv/dt or v*dv/dx methods

Suraiya, whose mass is \(m\) kg, takes a running jump into a swimming pool so that she begins to swim in a straight line with speed 0·2 ms\(^{-1}\). She continues to move in the same straight line, the only force acting on her being a resistance of magnitude \(mv^2 \sin \left(\frac{t}{100}\right)\) N, where \(v\) ms\(^{-1}\) is her speed at time \(t\) seconds after entering the pool and \(0 \leq t \leq 50\pi\).

1. Find an expression for \(v\) in terms of \(t\). [7 marks]
2. Calculate her greatest and least speeds during her motion. [3 marks]

A motor-cycle, whose mass including the rider is \(120\) kg, is decelerating on a horizontal straight road. The motor-cycle passes a point \(A\) with speed \(40 \text{ m s}^{-1}\) and when it has travelled a distance of \(x\) m beyond \(A\) its speed is \(v \text{ m s}^{-1}\). The engine develops a constant power of \(8\) kW and resistances are modelled by a force of \(0.25v^2\) N opposing the motion.

1. Show that \(\frac{480v^2}{v^3 - 32000} \frac{dv}{dx} = -1\). [5]
2. Find the speed of the motor-cycle when it has travelled \(500\) m beyond \(A\). [6]

\(O\) is a fixed point on a horizontal plane. A particle \(P\) of mass \(0.25\) kg is released from rest at \(O\) and moves in a straight line on the plane. At time \(t\) s after release the only horizontal force acting on \(P\) has magnitude $$\frac{1}{2400}(144 - t^2) \text{ N} \quad \text{for } 0 \leqslant t \leqslant 12$$ and $$\frac{1}{2400}(t^2 - 144) \text{ N} \quad \text{for } t \geqslant 12.$$ The force acts in the direction of \(P\)'s motion. \(P\)'s velocity at time \(t\) s is \(v\) m s\(^{-1}\).

1. Find an expression for \(v\) in terms of \(t\), valid for \(t \geqslant 12\), and hence show that \(v\) is three times greater when \(t = 24\) than it is when \(t = 12\). [8]
2. Sketch the \((t, v)\) graph for \(0 \leqslant t \leqslant 24\). [3]

A particle \(P\) of mass \(0.25\) kg is projected horizontally with speed \(5\) m s\(^{-1}\) from a fixed point \(O\) on a smooth horizontal surface and moves in a straight line on the surface. The only horizontal force acting on \(P\) has magnitude \(0.2v^2\) N, where \(v\) m s\(^{-1}\) is the velocity of \(P\) at time \(t\) s after it is projected from \(O\). This force is directed towards \(O\).

1. Find an expression for \(v\) in terms of \(t\). [5]

The particle \(P\) passes through a point \(X\) with speed \(0.2\) m s\(^{-1}\).

1. Find the average speed of \(P\) for its motion between \(O\) and \(X\). [5]

A particle of mass \(0.4\) kg, moving on a smooth horizontal surface, passes through a point \(O\) with velocity \(10\text{ ms}^{-1}\). At time \(t\) s after the particle passes through \(O\), the particle has a displacement \(x\) m from \(O\), has a velocity \(v\text{ ms}^{-1}\) away from \(O\), and is acted on by a force of magnitude \(\frac{1}{5}v\) N acting towards \(O\). Find

1. the time taken for the velocity of the particle to reduce from \(10\text{ ms}^{-1}\) to \(5\text{ ms}^{-1}\), [5]
2. the average velocity of the particle over this time. [6]

A particle \(Q\) of mass \(0.2\) kg is projected horizontally with velocity \(4\) m s\(^{-1}\) from a fixed point \(A\) on a smooth horizontal surface. At time \(t\) s after projection \(Q\) is \(x\) m from \(A\) and is moving away from \(A\) with velocity \(v\) m s\(^{-1}\). There is a force of \(3\cos 2t\) N acting on \(Q\) in the positive \(x\)-direction.

1. Find an expression for the velocity of \(Q\) at time \(t\). State the maximum and minimum values of the velocity of \(Q\) as \(t\) varies. [4]
2. Find the average velocity of \(Q\) between times \(t = \pi\) and \(t = \frac{3}{2}\pi\). [4]

A ball of mass \(m\) is thrown vertically upwards from the ground with an initial speed \(u\). When the speed of the ball is \(v\), the magnitude of the air resistance is \(mkv\), where \(k\) is a positive constant. By modelling the ball as a particle, find, in terms of \(u\), \(k\) and \(g\), the time taken for the ball to reach its greatest height. [8]

A small pebble of mass \(m\) is placed in a viscous liquid and sinks vertically from rest through the liquid. When the speed of the pebble is \(v\) the magnitude of the resistance due to the liquid is modelled as \(mkv^2\), where \(k\) is a positive constant. Find the speed of the pebble after it has fallen a distance \(D\) through the liquid. [11]

A particle \(P\) of mass 3 kg moves in a straight line on a smooth horizontal plane. When the speed of \(P\) is \(v\) m s\(^{-1}\), the resultant force acting on \(P\) is a resistance to motion of magnitude \(2v\) N. Find the distance moved by \(P\) while slowing down from 5 m s\(^{-1}\) to 2 m s\(^{-1}\). [5]

A car of mass \(M\) moves along a straight horizontal road. The total resistance to motion of the car is modelled as having constant magnitude \(R\). The engine of the car works at a constant rate \(RU\). Find the time taken for the car to accelerate from a speed of \(\frac{1}{4}U\) to a speed of \(\frac{1}{2}U\). [9]

A particle \(P\) of mass \(0.5\) kg is released from rest at time \(t = 0\) and falls vertically through a liquid. The motion of \(P\) is resisted by a force of magnitude \(2v\) N, where \(v\) m s\(^{-1}\) is the speed of \(v\) at time \(t\) seconds.

1. Show that \(5 \frac{\mathrm{d}v}{\mathrm{d}t} = 49 - 20v\). [2]
2. Find the speed of \(P\) when \(t = 1\). [5]

A lorry of mass \(M\) is moving along a straight horizontal road. The engine produces a constant driving force of magnitude \(F\). The total resistance to motion is modelled as having magnitude \(kv^2\), where \(k\) is a constant, and \(v\) is the speed of the lorry. Given the lorry moves with constant speed \(V\),

1. show that \(V = \sqrt{\frac{F}{k}}\). [2]

Given instead that the lorry starts from rest,

1. show that the distance travelled by the lorry in attaining a speed of \(\frac{1}{2}V\) is $$\frac{M}{2k}\ln\left(\frac{4}{3}\right).$$ [9]

A particle \(P\) of mass \(0.5\) kg falls vertically from rest. After \(t\) seconds it has speed \(v\) m s\(^{-1}\). A resisting force of magnitude \(1.5v\) newtons acts on \(P\) as it falls.

1. Show that \(3v = 9.8(1 - e^{-3t})\). [8]
2. Find the distance that \(P\) falls in the first two seconds of its motion. [5]

A small ball of mass \(m\) is projected vertically upwards from a point \(O\) with speed \(U\). The ball is subject to air resistance of magnitude \(mkv\), where \(v\) is the speed of the ball and \(k\) is a positive constant. Find, in terms of \(U\), \(g\) and \(k\), the maximum height above \(O\) reached by the ball. (8)

A car of mass 1000 kg is moving along a straight horizontal road. The engine of the car is working at a constant rate of 25 kW. When the speed of the car is \(v\) m s\(^{-1}\), the resistance to motion has magnitude \(10v\) newtons.

1. Show that, at the instant when \(v = 20\), the acceleration of the car is 1.05 m s\(^{-2}\). [3]
2. Find the distance travelled by the car as it accelerates from a speed of 10 m s\(^{-1}\) to a speed of 20 m s\(^{-1}\). [8]

A particle \(P\) of mass 2 kg moves in a straight line along a smooth horizontal plane. The only horizontal force acting on \(P\) is a resistance of magnitude \(4v\) N, where \(v\) m s\(^{-1}\) is its speed. At time \(t = 0\) s, \(P\) has a speed of 5 m s\(^{-1}\). Find \(v\) in terms of \(t\). [6]

A ball of mass \(m\) is thrown vertically upwards from the ground. When its speed is \(v\) the magnitude of the air resistance is modelled as being \(mkv^2\), where \(k\) is a positive constant. The ball is projected with speed \(\sqrt{\frac{g}{k}}\). By modelling the ball as a particle,

1. find the greatest height reached by the ball. [9]
2. State one physical factor which is ignored in this model. [1]

A rocket, with initial mass 1500 kg, including 600 kg of fuel, is launched vertically upwards from rest. The rocket burns fuel at a rate of 15 kg s\(^{-1}\) and the burnt fuel is ejected vertically downwards with a speed of 1000 m s\(^{-1}\) relative to the rocket. At time \(t\) seconds after launch \((t \leqslant 40)\) the rocket has mass \(m\) kg and velocity \(v\) m s\(^{-1}\).

1. Show that $$\frac{dv}{dt} + \frac{1000}{m} \frac{dm}{dt} = -9.8$$ [5]

1. Find \(v\) at time \(t\), \(0 \leqslant t \leqslant 40\) [5]

A space-ship is moving in a straight line in deep space and needs to reduce its speed from \(U\) to \(V\). This is done by ejecting fuel from the front of the space-ship at a constant speed \(k\) relative to the space-ship. When the speed of the space-ship is \(v\), its mass is \(m\).

1. Show that, while the space-ship is ejecting fuel, \(\frac{\mathrm{d}m}{\mathrm{d}v} = -\frac{m}{k}\). [6]

The initial mass of the space-ship is \(M\).

1. Find, in terms of \(U\), \(V\), \(k\) and \(M\), the amount of fuel which needs to be used to reduce the speed of the space-ship from \(U\) to \(V\). [6]

A rocket, with initial mass 1500 kg, including 600 kg of fuel, is launched vertically upwards from rest. The rocket burns fuel at a rate of 15 kg s\(^{-1}\) and the burnt fuel is ejected vertically downwards with a speed of 1000 m s\(^{-1}\) relative to the rocket. At time \(t\) seconds after launch \((t \leqslant 40)\) the rocket has mass \(m\) kg and velocity \(v\) m s\(^{-1}\).

1. Show that $$\frac{dv}{dt} + \frac{1000}{m}\frac{dm}{dt} = -9.8$$ [5]
2. Find \(v\) at time \(t\), \(0 \leqslant t \leqslant 40\) [5]

A spacecraft is travelling in a straight line in deep space where all external forces can be assumed to be negligible. The spacecraft decelerates by ejecting fuel at a constant speed \(k\) relative to the spacecraft, in the direction of motion of the spacecraft. At time \(t\), the spacecraft has speed \(v\) and mass \(m\).

1. Show, from first principles, that while the spacecraft is ejecting fuel, $$\frac{dv}{dm} - \frac{k}{m} = 0$$ [5]

At time \(t = 0\), the spacecraft has speed \(U\) and mass \(M\).

1. Find the mass of the spacecraft when it comes to rest. [6]

Given that \(m = Me^{-\alpha t^2}\), where \(\alpha\) is a positive constant, and that the spacecraft comes to rest at time \(t = T\),

1. find, in terms of \(U\) and \(T\) only, the distance travelled by the spacecraft in decelerating from speed \(U\) to rest. [6]

As a hailstone falls under gravity in still air, its mass increases. At time \(t\) the mass of the hailstone is \(m\). The hailstone is modelled as a uniform sphere of radius \(r\) such that $$\frac{dr}{dt} = kr,$$ where \(k\) is a positive constant.

1. Show that \(\frac{dm}{dt} = 3km\). [2]

Assuming that there is no air resistance,

1. show that the speed \(v\) of the hailstone at time \(t\) satisfies $$\frac{dv}{dt} = g - 3kv.$$ [4]

Given that the speed of the hailstone at time \(t = 0\) is \(u\),

1. find an expression for \(v\) in terms of \(t\). [5]
2. Hence show that the speed of the hailstone approaches the limiting value \(\frac{g}{3k}\). [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]

A particle of mass 4 kg moves horizontally in a straight line. At time \(t\) seconds the velocity of the particle is \(v\) m s\(^{-1}\) The following horizontal forces act on the particle: • a constant driving force of magnitude 1.8 newtons • another driving force of magnitude \(30\sqrt{t}\) newtons • a resistive force of magnitude \(0.08v^2\) newtons When \(t = 70\), \(v = 54\) Use Euler's method with a step length of 0.5 to estimate the velocity of the particle after 71 seconds. Give your answer to four significant figures. [6 marks]

A car of mass 1250 kg experiences a resistance to its motion of magnitude \(kv^2\) N, where \(k\) is a constant and \(v \, \text{m s}^{-1}\) is the car's speed. The car travels in a straight line along a horizontal road with its engine working at a constant rate of \(P\) W. At a point \(A\) on the road the car's speed is \(15 \, \text{m s}^{-1}\) and it has an acceleration of magnitude \(0.54 \, \text{m s}^{-2}\). At a point \(B\) on the road the car's speed is \(20 \, \text{m s}^{-1}\) and it has an acceleration of magnitude \(0.3 \, \text{m s}^{-2}\).

1. Find the values of \(k\) and \(P\). [7]

The power is increased to 15 kW.

1. Calculate the maximum steady speed of the car on a straight horizontal road. [3]