Ramiro and Lautaro are travelling from Buenos Aires to El Moro.

Ramiro travels in a vehicle whose velocity in \({\text{m}}{{\text{s}}^{ - 1}}\) is given by \({V_R} = 40 - {t^2}\), where \(t\) is in seconds.

Lautaro travels in a vehicle whose displacement from Buenos Aires in metres is given by \({S_L} = 2{t^2} + 60\).

When \(t = 0\), both vehicles are at the same point.

Find Ramiro’s displacement from Buenos Aires when \(t = 10\).

METHOD 1

\({S_L}(0) = 60\)   (seen anywhere)     (A1)

recognizing need to integrate \({V_R}\)     (M1)

eg     \({S_R}(t)\int {{V_R}{\text{d}}t} \)

correct expression     A1A1

eg     \(40t - \frac{1}{3}{t^3} + C\)

 

Note:     Award A1 for \(40t\), and A1 for \( - \frac{1}{3}{t^3}\).

 

equate displacements to find C     (R1)

eg     \(40(0) - \frac{1}{3}{(0)^3} + C = 60,{\text{ }}{S_L}(0) = {S_R}(0)\)

\(C = 60\)     A1

attempt to find displacement     (M1)

eg     \({S_R}(10),{\text{ }}40(10) - \frac{1}{3}{(10)^3} + 60\)

\(126.666\)

\(126\frac{2}{3}{\text{ (exact), 127 (m)}}\)     A1     N5

 

METHOD 2

recognizing need to integrate \({V_R}\)     (M1)

eg     \({S_R}(t) = \int {{V_R}{\text{d}}t} \)

valid approach involving a definite integral     (M1)

eg     \(\int_a^b {{V_R}{\text{d}}t} \)

correct expression with limits     (A1)

eg     \(\int_0^{10} {\left( {40 - {t^2}} \right){\text{d}}t,{\text{ }}} \int_0^{10} {{V_R}{\text{d}}t,{\text{ }}\left[ {40t - \frac{1}{3}{t^3}} \right]} _0^{10}\)

\(66.6666\)     A2

\({S_L}(0) = 60\)   (seen anywhere)     (A1)

valid approach to find total displacement     (M1)

eg     \(60 + 66.666\)

\(126.666\)

\(126\frac{2}{3}\)   (exact), \(127\) (m)     A1     N5

 

METHOD 3

\({S_L}(0) = 60\)     (seen anywhere)     (A1)

recognizing need to integrate \({V_R}\)     (M1)

eg     \({S_R}(t) = \int {{V_R}{\text{d}}t} \)

correct expression     A1A1

eg     \(40t - \frac{1}{3}{t^3} + C\)

 

Note:     Award A1 for \(40t\), and A1 for \( - \frac{1}{3}{t^3}\).

 

correct expression for Ramiro displacement     A1

eg     \({S_R}(10) - {S_R}(0),{\text{ }}\left[ {40t - \frac{1}{3}{t^3} + C} \right]_0^{10}\)

\(66.6666\)     A1

valid approach to find total displacement     (M1)

eg     \(60 + 66.6666\)

\(126\frac{2}{3}\) (exact), 127 (m)     A1     N5

[8 marks]

[N/A]