Given \(a = \frac{dv}{dt} = -(1 + v)\), rewrite the differential equation:

\(\frac{dv}{dt} + v = -1\) M1

This is a first-order linear differential equation. The integrating factor is:

\(e^{\int 1 \, dt} = e^t\) M1

Multiply through by \(e^t\):

\(\frac{d}{dt}(v e^t) = -e^t\) A1

Integrate both sides:

\(v e^t = -\int e^t \, dt = -e^t + c\) A1

\(v = -1 + c e^{-t}\)

Apply initial condition \(v = v_0\) at \(t = 0\):

\(v_0 = -1 + c \implies c = 1 + v_0\) A1

Thus, \(v(t) = (1 + v_0) e^{-t} – 1\) A1

[6 marks]

(i) At maximum displacement \(s_{\text{max}}\), \(v(T) = 0\). Using \(v(t) = (1 + v_0) e^{-t} – 1\):

\(v(T) = (1 + v_0) e^{-T} – 1 = 0\) M1

\((1 + v_0) e^{-T} = 1 \implies 1 + v_0 = e^T\) A1A1

[3 marks]

(ii) Since \(\frac{ds}{dt} = v(t) = (1 + v_0) e^{-t} – 1\), integrate to find \(s(t)\):

\(s = \int ((1 + v_0) e^{-t} – 1) \, dt = -(1 + v_0) e^{-t} – t + c\) M1A1

At \(t = 0\), \(s = 0\):

\(0 = -(1 + v_0) e^0 + c = -(1 + v_0) + c \implies c = 1 + v_0\)

\(s(t) = (1 + v_0) – (1 + v_0) e^{-t} – t\) A1

At \(t = T\), \(s = s_{\text{max}}\), and from (b)(i), \(e^T = 1 + v_0 \implies e^{-T} = \frac{1}{1 + v_0}\):

\(s_{\text{max}} = (1 + v_0) – (1 + v_0) e^{-T} – T = (1 + v_0) – (1 + v_0) \cdot \frac{1}{1 + v_0} – \ln(1 + v_0) = (1 + v_0) – 1 – \ln(1 + v_0) = v_0 – \ln(1 + v_0)\) A1

[4 marks]

Given \(v(T – k) = (1 + v_0) e^{-(T – k)} – 1\), and from (b)(i), \(1 + v_0 = e^T\):

\(v(T – k) = (1 + v_0) e^{-T} e^k – 1 = e^T \cdot e^{-T} e^k – 1 = e^k – 1\) M1A1

[2 marks]

For \(v(T + k) = (1 + v_0) e^{-(T + k)} – 1\):

Using \(1 + v_0 = e^T\):

\(v(T + k) = (1 + v_0) e^{-T} e^{-k} – 1 = e^T \cdot e^{-T} e^{-k} – 1 = e^{-k} – 1\) M1A1

[2 marks]

Using results from (c) and (d):

\(v(T – k) + v(T + k) = (e^k – 1) + (e^{-k} – 1) = e^k + e^{-k} – 2\) M1

Rewrite: \(e^k + e^{-k} – 2 = \frac{e^{2k} + 1 – 2e^k}{e^k} = \frac{(e^k – 1)^2}{e^k}\) A1

Since \((e^k – 1)^2 \geq 0\) and \(e^k > 0\), \(\frac{(e^k – 1)^2}{e^k} \geq 0\). Thus, \(v(T – k) + v(T + k) \geq 0\). R1

[3 marks]

Total [20 marks]