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Question
The acceleration, \(a{\text{ m}}{{\text{s}}^{ – 2}}\), of a particle at time t seconds is given by \[a = \frac{1}{t} + 3\sin 2t {\text{, for }} t \ge 1.\]
The particle is at rest when \(t = 1\) .
Find the velocity of the particle when \(t = 5\) .
Answer/Explanation
Markscheme
evidence of integrating the acceleration function (M1)
e.g. \(\int {\left( {\frac{1}{t} + 3\sin 2t} \right)} {\rm{d}}t\)
correct expression \(\ln t – \frac{3}{2}\cos 2t + c\) A1A1
evidence of substituting (1, 0) (M1)
e.g. \(0 = \ln 1 – \frac{3}{2}\cos 2 + c\)
\(c = – 0.624\) \(\left( { = \frac{3}{2}\cos 2 – \ln {\text{1 or }}\frac{{\rm{3}}}{{\rm{2}}}\cos 2} \right)\) (A1)
\(v = \ln t – \frac{3}{2}\cos 2t – 0.624\) \(\left( { = \ln t – \frac{3}{2}\cos 2t + \frac{3}{2}\cos {\text{2 or ln}}t – \frac{3}{2}\cos 2t + \frac{3}{2}\cos 2 – \ln 1} \right)\) (A1)
\(v(5) = 2.24\) (accept the exact answer \(\ln 5 – 1.5\cos 10 + 1.5\cos 2\) ) A1 N3
[7 marks]
Question
A gradient function is given by \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 10{{\rm{e}}^{2x}} – 5\) . When \(x = 0\) , \(y = 8\) . Find the value of y when \(x = 1\) .
Answer/Explanation
Markscheme
METHOD 1
evidence of anti-differentiation (M1)
e.g. \(\int {(10{{\rm{e}}^{2x}} – 5){\rm{d}}x} \)
\(y = 5{{\rm{e}}^{2x}} – 5x + C\) A2A1
Note: Award A2 for \(5{{\rm{e}}^{2x}}\) , A1 for \( – 5x\) . If “C” is omitted, award no further marks.
substituting \((0{\text{, }}8)\) (M1)
e.g. \(8 = 5 + C\)
\(C = 3\) \((y = 5{{\rm{e}}^{2x}} – 5x + 3)\) (A1)
substituting \(x = 1\) (M1)
\(y = 34.9\) \((5{{\rm{e}}^2} – 2)\) A1 N4
METHOD 2
evidence of definite integral function expression (M2)
e.g. \(\int_a^x {f'(t){\rm{d}}t = } f(x) – f(a)\) , \(\int_0^x {(10{{\rm{e}}^{2x}} – 5)} \)
initial condition in definite integral function expression (A2)
e.g. \(\int_0^x {(10{{\rm{e}}^{2t}} – 5)} {\rm{d}}t = y – 8\) , \(\int_0^x {(10{{\rm{e}}^{2x}} – 5)} {\rm{d}}x + 8\)
correct definite integral expression for y when \(x = 1\) (A2)
e.g. \(\int_0^1 {(10{{\rm{e}}^{2x}} – 5){\rm{d}}x + 8} \)
\(y = 34.9\) \((5{{\rm{e}}^2} – 2)\) A1 N4
[8 marks]