5. A car of mass 1000 kg is moving along a straight horizontal road. The engine of the car produces a constant driving force of 800 N. At time \(t\) seconds, \(t \geqslant 0\), the speed of the car is \(v\ \text{m s}^{-1}\) and the resistance to the motion of the car has magnitude \(2v^2\) N.
The time taken for the speed of the car to increase from \(5\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) is \(T\) seconds.
(a) Show that \(T = \dfrac{25}{2}\ln\left(\dfrac{21}{5}\right)\) (7)
When \(t = 0\), \(v = 2\)
(b) Show that \(\dfrac{20 + v}{20 - v} = p\mathrm{e}^{qt}\) where \(p\) and \(q\) are rational numbers to be found. (2)
(c) Hence explain why \(v \lt 20\) for all values of \(t \geqslant 0\) (1)
M1: Form equation of motion. Need all terms and dimensionally correct. Condone any correct form for acceleration and sign errors
A1: Unsimplified equation in \(v\) and \(t\) with at most one error
A1: Correct unsimplified equation in \(v\) and \(t\)
M1: Separate variables and integrate to a correct logarithmic form. May see attempts using substitution / trig etc. but need to get to logarithmic form.
A1: Any equivalent form. Constant of integration and modulus signs not required.
M1: Use correct limits correctly in an expression containing \(\lambda\ln(20 - v)\) and \(\mu\ln(20 + v)\)
A1*: Obtain given answer from correct working (condone disappearance of “\(\mathrm{d}v\)” or “\(\mathrm{d}t\)” on integrals)
Mark scheme (b)
Scheme
Marks
AO
\(t = 0,\ v = 2\ \Rightarrow\ C = \dfrac{25}{2}\ln\dfrac{22}{18}\ \left(= \dfrac{25}{2}\ln\dfrac{11}{9}\right)\)
M1: Use boundary condition to find constant of integration for their integral.
A1*: Obtain given conclusion from correct working. Allow any exact equivalent for \(p\) & \(q\).
Mark scheme (c)
Scheme
Marks
AO
As \(\mathrm{e}^{qt} \gt 0\), \(v \lt 20\) *
B1ft *
2.4
(1)
(10 marks)
Notes
B1*: Obtain given conclusion. Follow their \(p \gt 0\) but must be referring to the idea that the expression must be positive because it is an exponential function.
2. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\ \text{m s}^{-1}\) in the positive \(x\)-direction, where
\[v = 3 - \sqrt{2t+1} \qquad t \geqslant 0\]
(a) Find the value of \(t\) when \(P\) is at instantaneous rest. (2)
The acceleration of \(P\) at time \(t\) seconds is \(a\ \text{m s}^{-2}\) in the positive \(x\)-direction.
(b) Show that \(a = \dfrac{1}{v-3}\) (3)
(c) Find the speed of \(P\) when it is decelerating at \(\dfrac{4}{3}\ \text{m s}^{-2}\) (2)
(d) Find the total distance travelled by \(P\) between \(t = 0\) and \(t = \dfrac{15}{2}\) [Solutions relying on calculator technology are not acceptable.] (4)
Mark scheme (a)
Scheme
Marks
AO
\(0 = 3 - \sqrt{2t+1}\)
M1
2.1
\(t = 4\)
A1
1.1b
(2)
Notes
M1: Correct equation
A1: cao
Mark scheme (b)
Scheme
Marks
AO
Differentiate \(v\) wrt \(t\)
M1
2.1
\(a = \dfrac{-1}{\sqrt{2t+1}}\)
A1
1.1b
\(a = \dfrac{1}{v-3}\) *
A1*
2.2a
(3)
Notes
M1: Both powers decreasing by 1
A1: Correct expression
A1*: Given answer correctly obtained including “\(a =\)” seen in solution
3. A particle \(P\) is moving along the \(x\)-axis. At time \(t\) seconds, \(P\) has velocity \(v\ \text{m s}^{-1}\) in the positive \(x\) direction and acceleration \(a\ \text{m s}^{-2}\) in the positive \(x\) direction.
In a model of the motion of \(P\)
\[a = 4 - 3v\]
When \(t = 0\), \(v = 0\)
(a) Use integration to show that \(v = k\left(1 - \mathrm{e}^{-3t}\right)\), where \(k\) is a constant to be found. (7)
When \(t = 0\), \(P\) is at the origin \(O\)
(b) Find, in terms of \(t\) only, the distance of \(P\) from \(O\) at time \(t\) seconds. (4)
M1: Form a differential equation in \(v\) and \(t\) and integrate. Must attempt integration of \(\dfrac{k}{(3t + 5)^3}\). RHS can be implied.
A1: Correct integration. Ignore any limits. Accept without constant of integration.
M1: Use \(v = 0,\ t = 0\) as limits in a definite integral or to find the constant of integration
A1*: Obtain given answer in the form \(v = p - \dfrac{q}{(3t + 5)^2}\) from correct working. Accept if correct form given and values of \(p\) and \(q\) stated separately. Must have “\(v =\)”.
Mark scheme (b)
Scheme
Marks
AO
\(t \to \infty \Rightarrow v \to \dfrac{16}{25}\ (= 0.64)\)
2. 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 \((t \gt 0)\),
the particle is moving with speed \(v\ \text{m s}^{-1}\)
\(F = 2 + v\)
The time taken for the speed of the particle to increase from \(5\ \text{m s}^{-1}\) to \(10\ \text{m s}^{-1}\) is \(T\) seconds.
(a) Show that \(T = 2\ln\dfrac{12}{7}\) (4)
The distance moved by the particle as its speed increases from \(5\ \text{m s}^{-1}\) to \(10\ \text{m s}^{-1}\) is \(D\) metres.
M1: Obtain a differential equation in \(v\) and \(x\) Allow \(\pm\) for the acceleration
M1: Separate variables and integrate. If using integration by parts they need to complete the integration to score M1.
A1: Or equivalent unsimplified form. Allow without modulus signs.
A1: Use limits to eliminate constant of integration, or in a definite integral, to obtain exact distance from exact working. Any equivalent simplified form.
2. A particle \(P\) is moving along the \(x\)-axis. At time \(t\) seconds, \(t \geqslant 0\), \(P\) has acceleration \(a\ \text{m s}^{-2}\) and velocity \(v\ \text{m s}^{-1}\) in the direction of \(x\) increasing, where
\[v = \mathrm{e}^{2t} + 6\mathrm{e}^{t} - kt\]
and \(k\) is a positive constant.
When \(t = \ln 2\), \(a = 0\)
(a) Find the value of \(k\). (4)
When \(t = 0\), the particle passes through the fixed point \(A\). When \(t = \ln 2\), the particle is \(d\) metres from \(A\).
(b) Showing all stages of your working, find the value of \(d\) correct to 2 significant figures. [Solutions relying entirely on calculator technology are not acceptable.] (4)
Mark scheme (a)
Scheme
Marks
AO
Use of \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t}\)
M1
3.1a
\(a = 2\mathrm{e}^{2t} + 6\mathrm{e}^{t} - k\)
A1
1.1b
Substitute \(t = \ln 2\) into their acceleration and solve for \(k\)
M1
1.1b
\(k = 20\)
A1
2.2a
(4)
Notes
M1: Use the model and differentiate \(v\) to obtain \(a\). Obtain form \(p\mathrm{e}^{2t} + q\mathrm{e}^{t} - k\)
A1: Correct only
M1: \((2 \times 4 + 6 \times 2 - k = 0)\) Their acceleration must come from an attempt to differentiate.
M1: Integrate \(v\) to obtain \(x\). Obtain form \(a\mathrm{e}^{2t} + b\mathrm{e}^{t} + ct^2\)
A1ft: Allow in \(k\) or their \(k\)
M1: Use the model to evaluate constant of integration or use boundary conditions as limits in a definite integral. Their \(x\) must come from an attempt to integrate.
4. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\ \text{m s}^{-1}\) in the direction of \(x\) increasing, where
\[v = \dfrac{1}{2}\left(3\mathrm{e}^{2t} - 1\right) \qquad t \geqslant 0\]
The acceleration of \(P\) at time \(t\) seconds is \(a\ \text{m s}^{-2}\)
(a) Show that \(a = 2v + 1\) (2)
(b) Find the acceleration of \(P\) when \(t = 0\) (1)
(c) Find the exact distance travelled by \(P\) in accelerating from a speed of \(1\ \text{m s}^{-1}\) to a speed of \(4\ \text{m s}^{-1}\) (7)
2. 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 \(v\ \text{m s}^{-1}\) and the magnitude of the resistance to motion is \(3v^2\) N.
(a) Show that \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{200 - 3v^3}{60v^2}\) (4)
The distance travelled by the cyclist as her speed increases from \(2\ \text{m s}^{-1}\) to \(4\ \text{m s}^{-1}\) is \(D\) metres.
(b) Find the exact value of \(D\) (3)
Mark scheme (a)
Scheme
Marks
AO
Use of \(P = Fv\)
B1
3.3
Equation of motion \(\left(F - 3v^2 = 60a\right)\)
B1: Seen or implied Not just quoted. Need at least \(200 = Fv\) Could be on its own, in an equation or on a diagram
M1: Form equation of motion. Need all terms and dimensionally correct. Condone any correct form for acceleration and sign errors Allow with \(m\) not substituted
A1: Correct equation - any equivalent form with correct acceleration
A1*: Obtain given answer from correct working Must be as written in the question but could swap LHS and RHS
M1: Separate variables and integrate to obtain \((x =)\,k\ln(\ldots\ldots)\) (Constant of integration not required) Condone if the \(x\) is not explicitly stated but M0 if it is an incorrect function.
M1: Use limits correctly in an expression containing \(k\ln\left(200 - 3v^3\right)\) to find \(D\) Substitute and subtract in the correct order
A1: Obtain exact answer from correct working Any equivalent single term No working seen is Max M1M0A0
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 \(kv^2\), where \(k\) is a constant.
(a) Show that \(t = \dfrac{V}{2g}\ln\left(\dfrac{V + v}{V - v}\right)\) where \(V^2 = \dfrac{mg}{k}\) (4)
(b) Give an interpretation of the value of \(V\), justifying your answer. (2)
At time \(t\), \(P\) has fallen a distance \(s\).
(c) Show that \(s = \dfrac{V^2}{2g}\ln\left(\dfrac{V^2}{V^2 - v^2}\right)\) (4)
Mark scheme (a)
Scheme
Marks
AO
\(mg - kv^2 = m\dfrac{\mathrm{d}v}{\mathrm{d}t}\)
M1
2.5
Separate variables and integrate
M1
2.1
A correct equation in any form (ignore constant or limits) e.g. \(t = \dfrac{m}{k}\,\dfrac{1}{2\sqrt{\frac{mg}{k}}}\ln\left(\dfrac{\sqrt{\frac{mg}{k}} + v}{\sqrt{\frac{mg}{k}} - v}\right) \quad (+\,C)\)
3. 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 \(v\ \text{m s}^{-1}\). The resultant force acting on \(P\) is directed towards \(O\) and has magnitude \(kv^2\) N, where \(k\) is a positive constant.
When \(x = 1\), \(v = 4\) and when \(x = 2\), \(v = 2\)
(a) Show that \(v = ab^x\), where \(a\) and \(b\) are constants to be found. (6)
The time taken for the speed of \(P\) to decrease from \(4\ \text{m s}^{-1}\) to \(2\ \text{m s}^{-1}\) is \(T\) seconds.
(b) Show that \(T = \dfrac{1}{4\ln 2}\) (4)
Mark scheme (a)
Scheme
Marks
AO
Form differential equation: \(0.5a = 0.5v\dfrac{\mathrm{d}v}{\mathrm{d}x} = -kv^2\)
3. 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 \((t \gt 0)\) after leaving \(O\), the velocity of the car is \(v\ \text{m s}^{-1}\) and the acceleration of the car is modelled as \((p + qv)\ \text{m s}^{-2}\), where \(p\) and \(q\) are constants.
When \(t = 0\), the acceleration of the car is \(3\ \text{m s}^{-2}\)
When \(t = T\), the acceleration of the car is \(\dfrac{1}{2}\ \text{m s}^{-2}\) and \(v = 4\)
(a) Show that \[8\frac{\mathrm{d}v}{\mathrm{d}t} = (24 - 5v)\] (6)
(b) Find the exact value of \(T\), simplifying your answer. (6)
2. 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, \(t \gt 0\), \(P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v\ \text{m s}^{-1}\). The force is acting in the direction of \(x\) increasing and has magnitude \(\dfrac{k}{v}\) newtons, where \(k\) is a constant.
At \(x = 3\), \(v = 2\) and at \(x = 6\), \(v = 2.5\)
(a) Show that \(v^3 = \dfrac{61x + 9}{24}\) (6)
The time taken for the speed of \(P\) to increase from \(2\ \text{m s}^{-1}\) to \(2.5\ \text{m s}^{-1}\) is \(T\) seconds.
(b) Use algebraic integration to show that \(T = \dfrac{81}{61}\) (4)
Mark scheme (a)
Scheme
Marks
AO
Strategy to find \(v^3\) in terms of \(x\)
M1
3.1a
Differential equation in \(v\) and \(x\): \(\ 0.4v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{k}{v}\)
2. A car moves in a straight line along a horizontal road. The car is modelled as a particle. At time \(t\) seconds, where \(t \geqslant 0\), the speed of the car is \(v\ \text{m s}^{-1}\)
At the instant when \(t = 0\), the car passes through the point \(A\) with speed \(2\ \text{m s}^{-1}\)
The acceleration, \(a\ \text{m s}^{-2}\), of the car is modelled by
\[a = \frac{4}{2+v}\]
in the direction of motion of the car.
(a) Use algebraic integration to show that \(v = \sqrt{8t+16} - 2\) (6)
At the instant when the car passes through the point \(B\), the speed of the car is \(4\ \text{m s}^{-1}\)
(b) Use algebraic integration to find the distance \(AB\). (6)
B1: Use the result from (a) to find \(t\) when \(v = 4\): seen or implied
M1: Form differential equation in \(x\) and \(t\)
M1: Integrate to obtain terms of the correct form. Condone missing constant of integration.
A1: Correct integration. Condone missing constant of integration.
M1: Use boundary conditions in the model to find constant of integration, or as limits on a definite integral. Note this is an independent M mark. M0 if they use \(t = 4\)
4. A particle, \(P\), moves on the \(x\)-axis. At time \(t\) seconds, \(t \geqslant 0\), the velocity of \(P\) is \(v\ \text{m s}^{-1}\) in the direction of \(x\) increasing and the acceleration of \(P\) is \(a\ \text{m s}^{-2}\) in the direction of \(x\) increasing.
When \(t = 0\) the particle is at rest at the origin \(O\).
Given that \(a = \dfrac{5}{2}(5 - v)\)
(a) show that \(v = 5\left(1 - \mathrm{e}^{-2.5t}\right)\) (5)
(b) state the limiting value of \(v\) as \(t\) increases. (1)
At the instant when \(v = 2.5\), the particle is \(d\) metres from \(O\).
