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

1. At time \(t = 0\), a toy electric car is at rest at a fixed point \(O\). The car then moves in a horizontal straight line so that at time \(t\) seconds \(( t > 0 )\) after leaving \(O\), the velocity of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the acceleration of the car is modelled as \(( p + q v ) \mathrm { ms } ^ { - 2 }\), where \(p\) and \(q\) are constants.

When \(t = 0\), the acceleration of the car is \(3 \mathrm {~ms} ^ { - 2 }\) When \(t = T\), the acceleration of the car is \(\frac { 1 } { 2 } \mathrm {~ms} ^ { - 2 }\) and \(v = 4\)

1. Show that $$8 \frac { \mathrm {~d} v } { \mathrm {~d} t } = ( 24 - 5 v )$$
2. Find the exact value of \(T\), simplifying your answer.

1. A car of mass 600 kg pulls a trailer of mass 150 kg along a straight horizontal road. The trailer is connected to the car by a light inextensible towbar, which is parallel to the direction of motion of the car. The resistance to the motion of the trailer is modelled as a constant force of magnitude 200 N . At the instant when the speed of the car is \(v \mathrm {~ms} ^ { - 1 }\), the resistance to the motion of the car is modelled as a force of magnitude \(( 200 + \lambda v ) \mathrm { N }\), where \(\lambda\) is a constant.

When the engine of the car is working at a constant rate of 15 kW , the car is moving at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)

1. Show that \(\lambda = 8\) Later on, the car is pulling the trailer up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 15 }\) The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 200 N at all times. At the instant when the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the car from non-gravitational forces is modelled as a force of magnitude \(( 200 + 8 v ) \mathrm { N }\). The engine of the car is again working at a constant rate of 15 kW .
   When \(v = 10\), the towbar breaks. The trailer comes to instantaneous rest after moving a distance \(d\) metres up the road from the point where the towbar broke.
2. Find the acceleration of the car immediately after the towbar breaks.
3. Use the work-energy principle to find the value of \(d\).
   

1. A truck of mass 1200 kg is moving along a straight horizontal road.

At the instant when the speed of the truck is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the truck is modelled as a force of magnitude \(( 900 + 9 v ) \mathrm { N }\). The engine of the truck is working at a constant rate of 25 kW .

1. Find the deceleration of the truck at the instant when \(v = 25\) Later on, the truck is moving up a straight road that is inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 20 }\) At the instant when the speed of the truck is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the truck from non-gravitational forces is modelled as a force of magnitude ( \(900 + 9 v\) ) N. When the engine of the truck is working at a constant rate of 25 kW the truck is moving up the road at a constant speed of \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
2. Find the value of \(V\).

1. A van of mass 900 kg is moving along a straight horizontal road.

At the instant when the speed of the van is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the van is modelled as a force of magnitude \(( 500 + 7 v ) \mathrm { N }\). When the engine of the van is working at a constant rate of 18 kW , the van is moving along the road at a constant speed \(V \mathrm {~ms} ^ { - 1 }\)

1. Find the value of \(V\). Later on, the van is moving up a straight road that is inclined to the horizontal at an angle \(\theta\), where \(\sin \theta = \frac { 1 } { 21 }\) At the instant when the speed of the van is \(v \mathrm {~ms} ^ { - 1 }\), the resistance to the motion of the van from non-gravitational forces is modelled as a force of magnitude \(( 500 + 7 v ) \mathrm { N }\). The engine of the van is again working at a constant rate of 18 kW .
2. Find the acceleration of the van at the instant when \(v = 15\)

1. A particle, \(P\), of mass 0.4 kg is moving along the positive \(x\)-axis, in the positive \(x\) direction under the action of a single force. At time \(t\) seconds, \(t > 0 , P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The force is acting in the direction of \(x\) increasing and has magnitude \(\frac { k } { v }\) newtons, where \(k\) is a constant.

At \(x = 3 , v = 2\) and at \(x = 6 , v = 2.5\)

1. Show that \(v ^ { 3 } = \frac { 61 x + 9 } { 24 }\) The time taken for the speed of \(P\) to increase from \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.
2. Use algebraic integration to show that \(T = \frac { 81 } { 61 }\)

1. A particle \(P\) of mass 0.5 kg is moving along the positive \(x\)-axis in the direction of \(x\) increasing. At time \(t\) seconds \(( t \geqslant 0 ) , P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The resultant force acting on \(P\) is directed towards \(O\) and has magnitude \(k v ^ { 2 } \mathrm {~N}\), where \(k\) is a positive constant.

When \(x = 1 , v = 4\) and when \(x = 2 , v = 2\)

1. Show that \(v = a b ^ { x }\), where \(a\) and \(b\) are constants to be found. The time taken for the speed of \(P\) to decrease from \(4 \mathrm {~ms} ^ { - 1 }\) to \(2 \mathrm {~ms} ^ { - 1 }\) is \(T\) seconds.
2. Show that \(T = \frac { 1 } { 4 \ln 2 }\)

At time \(t = 0\), a small stone \(P\) of mass \(m\) is released from rest and falls vertically through the air. At time \(t\), the speed of \(P\) is \(v\) and the resistance to the motion of \(P\) from the air is modelled as a force of magnitude \(k v ^ { 2 }\), where \(k\) is a constant.

1. Show that \(t = \frac { V } { 2 g } \ln \left( \frac { V + v } { V - v } \right)\) where \(V ^ { 2 } = \frac { m g } { k }\)
2. Give an interpretation of the value of \(V\), justifying your answer. At time \(t , P\) has fallen a distance \(s\).
3. Show that \(s = \frac { V ^ { 2 } } { 2 g } \ln \left( \frac { V ^ { 2 } } { V ^ { 2 } - v ^ { 2 } } \right)\)

1. A cyclist and her cycle have a combined mass of 60 kg . The cyclist is moving along a straight horizontal road and is working at a constant rate of 200 W .

When she has travelled a distance \(x\) metres, her speed is \(v \mathrm {~ms} ^ { - 1 }\) and the magnitude of the resistance to motion is \(3 v ^ { 2 } \mathrm {~N}\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} x } = \frac { 200 - 3 v ^ { 3 } } { 60 v ^ { 2 } }\) The distance travelled by the cyclist as her speed increases from \(2 \mathrm {~ms} ^ { - 1 }\) to \(4 \mathrm {~ms} ^ { - 1 }\) is \(D\) metres.
2. Find the exact value of \(D\)

1. A particle of mass 2 kg is moving in a straight line on a smooth horizontal surface under the action of a horizontal force of magnitude \(F\) newtons.

At time \(t\) seconds \(( t > 0 )\),

• the particle is moving with speed \(v \mathrm {~ms} ^ { - 1 }\)
• \(F = 2 + v\)

The time taken for the speed of the particle to increase from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.

1. Show that \(T = 2 \ln \frac { 12 } { 7 }\) The distance moved by the particle as its speed increases from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~ms} ^ { - 1 }\) is \(D\) metres.
2. Find the exact value of \(D\).

1. A car of mass 500 kg moves along a straight horizontal road.

The engine of the car produces a constant driving force of 1800 N .
The car accelerates from rest from the fixed point \(O\) at time \(t = 0\) and at time \(t\) seconds the car is \(x\) metres from \(O\), moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the car has magnitude \(2 v ^ { 2 } \mathrm {~N}\). At time \(T\) seconds, the car is at the point \(A\), moving with speed \(10 \mathrm {~ms} ^ { - 1 }\).

1. Show that \(T = \frac { 25 } { 6 } \ln 2\)
2. Show that the distance from \(O\) to \(A\) is \(125 \ln \frac { 9 } { 8 } \mathrm {~m}\).

8 A piston of mass 1.5 kg moves in a straight line inside a long straight horizontal cylinder. At time \(t \mathrm {~s}\) the displacement of the piston from its initial position at one end of the cylinder is \(x \mathrm {~m}\) and its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{a8c9d007-e67f-4637-9e74-630ba9a91442-5_168_805_1726_630} The piston starts moving when \(t = 2\) and is brought to rest when it reaches the other end of the cylinder. While the piston is in motion it is acted on by a force of magnitude \(\frac { 6 } { t ^ { 2 } } \mathrm {~N}\) in the positive \(x\) direction, and also by a force of magnitude \(\frac { 3 v } { t } \mathrm {~N}\) resisting the motion.

1. Show that, while the piston is in motion, \(\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 2 v } { t } = \frac { 4 } { t ^ { 2 } }\). The piston reaches the other end of the cylinder when \(t = 20\).
2. Find the speed of the piston immediately before it is brought to rest.
3. Show that the piston travels a distance of 5.61 m , correct to 3 significant figures. \section*{OCR} \section*{Oxford Cambridge and RSA}

4 A particle \(P\) of mass 8 kg moves in a straight line on a smooth horizontal plane. At time \(t \mathrm {~s}\) the displacement of \(P\) from a fixed point \(O\) on the line is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). Initially, \(P\) is at rest at \(O\). \(P\) is acted on by a horizontal force, directed along the line away from \(O\), with magnitude proportional to \(\sqrt { 9 + v ^ { 2 } }\). When \(v = 1.25\), the magnitude of this force is 13 N .

1. Show that \(\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
3. Find an expression for \(x\) in terms of \(t\) for \(t \geqslant 0\).

3 A body, \(Q\), of mass 2 kg moves in a straight line under the action of a single force which acts in the direction of motion of \(Q\). Initially the speed of \(Q\) is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). At time \(t \mathrm {~s}\), the magnitude \(F N\) of the force is given by $$F = t ^ { 2 } + 3 \mathrm { e } ^ { t } , \quad 0 \leq t \leq 4 .$$

1. Calculate the impulse of the force over the time interval.
2. Hence find the speed of \(Q\) when \(t = 4\).

5 A car, of mass 1600 kg , is travelling along a straight horizontal road at a speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when the driving force is removed. The car then freewheels and experiences a resistance force. The resistance force has magnitude \(40 v\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the car after it has been freewheeling for \(t\) seconds. Find an expression for \(v\) in terms of \(t\).

7 A motorcycle has a maximum power of 72 kilowatts. The motorcycle and its rider are travelling along a straight horizontal road. When they are moving at a speed of \(\mathrm { V } \mathrm { m } \mathrm { s } ^ { - 1 }\), they experience a total resistance force of magnitude \(k V\) newtons, where \(k\) is a constant.

1. The maximum speed of the motorcycle and its rider is \(60 \mathrm {~ms} ^ { - 1 }\). Show that \(k = 20\).
2. When the motorcycle is travelling at \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the rider allows the motorcycle to freewheel so that the only horizontal force acting is the resistance force. When the motorcycle has been freewheeling for \(t\) seconds, its speed is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the magnitude of the resistance force is \(20 v\) newtons. The mass of the motorcycle and its rider is 500 kg .
   1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - \frac { v } { 25 }\).
   2. Hence find the time that it takes for the speed of the motorcycle to reduce from \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
      (6 marks)

8 A stone, of mass 0.05 kg , is moving along the smooth horizontal floor of a tank, which is filled with oil. At time \(t\), the stone has speed \(v\). As the stone moves, it experiences a resistance force of magnitude \(0.08 v ^ { 2 }\).

1. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 1.6 v ^ { 2 }$$ (2 marks)
2. The initial speed of the stone is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Show that $$v = \frac { 15 } { 5 + 24 t }$$ (5 marks)

5 A golf ball, of mass \(m \mathrm {~kg}\), is moving in a straight line across smooth horizontal ground. At time \(t\) seconds, the golf ball has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). As the golf ball moves, it experiences a resistance force of magnitude \(0.2 m v ^ { \frac { 1 } { 2 } }\) newtons until it comes to rest. No other horizontal force acts on the golf ball. Model the golf ball as a particle.

1. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.2 v ^ { \frac { 1 } { 2 } }$$
2. When \(t = 0\), the speed of the golf ball is \(16 \mathrm {~ms} ^ { - 1 }\). Show that \(v = ( 4 - 0.1 t ) ^ { 2 }\).
3. Find the value of \(t\) when \(v = 1\).
4. Find the distance travelled by the golf ball as its speed decreases from \(16 \mathrm {~ms} ^ { - 1 }\) to \(1 \mathrm {~ms} ^ { - 1 }\).

6 A car, of mass \(m\), is moving along a straight smooth horizontal road. At time \(t\), the car has speed \(v\). As the car moves, it experiences a resistance force of magnitude \(0.05 m v\). No other horizontal force acts on the car.

1. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.05 v$$
2. When \(t = 0\), the speed of the car is \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Show that \(v = 20 \mathrm { e } ^ { - 0.05 t }\).
3. Find the time taken for the speed of the car to reduce to \(10 \mathrm {~ms} ^ { - 1 }\).

8 A stone, of mass \(m\), is moving in a straight line along smooth horizontal ground.
At time \(t\), the stone has speed \(v\). As the stone moves, it experiences a total resistance force of magnitude \(\lambda m v ^ { \frac { 3 } { 2 } }\), where \(\lambda\) is a constant. No other horizontal force acts on the stone.

1. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - \lambda v ^ { \frac { 3 } { 2 } }$$ (2 marks)
2. The initial speed of the stone is \(9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Show that $$v = \frac { 36 } { ( 2 + 3 \lambda t ) ^ { 2 } }$$ (7 marks)
3. Find, in terms of \(\lambda\), the time taken for the speed of the stone to drop to \(4 \mathrm {~ms} ^ { - 1 }\).

2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.

3 A particle \(P\) of mass 8 kg moves in a straight line on a smooth horizontal plane. At time \(t \mathrm {~s}\) the displacement of \(P\) from a fixed point \(O\) on the line is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). Initially, \(P\) is at rest at \(O\). \(P\) is acted on by a horizontal force, directed along the line away from \(O\), with magnitude proportional to \(\sqrt { 9 + v ^ { 2 } }\). When \(v = 1.25\), the magnitude of this force is 13 N .

1. Show that \(\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
3. Find an expression for \(x\) in terms of \(t\) for \(t \geqslant 0\).

An object, of mass 1.8 kg , is falling vertically downwards under gravity. During the motion, it experiences a variable resistance of \(0 \cdot 2 v ^ { 2 } \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the object at time \(t\) seconds.

1. Show that \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 9 g - v ^ { 2 } } { 9 }$$ At time \(t = 0\), the object passes a point \(A\) with a speed of \(\sqrt { g } \mathrm {~ms} ^ { - 1 }\). The object then hits the ground with a speed of \(8 \mathrm {~ms} ^ { - 1 }\).
2. Calculate the time taken for the object to hit the ground.
   …………………………………………………………………………………………………………………………………………………... \includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_67_1614_639_264} \includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_76_1614_717_264} \includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_79_1614_801_264} \includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_72_1609_895_267} \includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_75_1614_979_264}
   
3. Given that the distance of the object from \(A\) at time \(t\) is \(x\) metres, form another differential equation to find an expression for \(x\) in terms of \(v\). Hence, find the height of \(A\) above the ground.
   \section*{PLEASE DO NOT WRITE ON THIS PAGE}

1 A lorry moves along a straight horizontal road. The engine of the lorry produces a constant power of 80 kW . The mass of the lorry is 10 tonnes and the resistance to motion is constant at 4000 N .

1. Express the driving force of the lorry in terms of its velocity and hence, using Newton's second law, write down a differential equation which connects the velocity of the lorry and the time for which it has been moving.
2. Hence find the time taken, in seconds, for the lorry to accelerate from \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

12 A train of mass 250 tonnes is ascending an incline of \(\sin ^ { - 1 } \left( \frac { 1 } { 500 } \right)\) and working at 400 kW against resistance to motion which may be regarded as a constant force of 20000 N .

1. Find the constant speed, \(V\), with which the train can ascend the incline working at this power.
2. The train begins to ascend the incline at \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at the same power and against the same resistance. Find the distance covered in reaching a speed of \(\frac { 3 } { 4 } V\).

7 A cyclist and her machine have a combined mass of 90 kg and she is riding along a straight horizontal road. She is working at a constant power of 75 W . At time \(t\) seconds her speed is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the resistance to motion is \(k v \mathrm {~N}\), where \(k\) is a constant.

1. If the cyclist's maximum steady speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), show that \(k = \frac { 3 } { 4 }\).
2. Use Newton's second law to show that $$\frac { 25 } { v } - \frac { v } { 4 } = 30 \frac { \mathrm {~d} v } { \mathrm {~d} t } .$$
3. Find the time taken for the cyclist to accelerate from a speed of \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to a speed of \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).