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Date May 2013 Marks available 7 Reference code 13M.1.sl.TZ2.6
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 6 Adapted from N/A

Question

A rocket moving in a straight line has velocity \(v\) km s–1 and displacement \(s\) km at time \(t\) seconds. The velocity \(v\) is given by \(v(t) = 6{{\rm{e}}^{2t}} + t\) . When \(t = 0\) , \(s = 10\) .

Find an expression for the displacement of the rocket in terms of \(t\) .

Markscheme

evidence of anti-differentiation     (M1)

eg   \(\int {(6{{\rm{e}}^{2t}} + t)} \)

\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + C\)     A2A1

Note: Award A2 for \(3{{\rm{e}}^{2t}}\) , A1 for \(\frac{{{t^2}}}{2}\) .

 

attempt to substitute (\(0\), \(10\)) into their integrated expression (even if \(C\) is missing)     (M1) 

correct working     (A1)

eg   \(10 = 3 + C\) , \(C = 7\)

\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7\)     A1     N6

Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of \({\rm{e}}\).

[7 marks]

Examiners report

A good number of candidates earned full marks on this question, and many others were able to earn at least half of the available marks. Most candidates knew to integrate, but there were quite a few who tried to find the derivative instead. Many candidates integrated the term \(6{{\text{e}}^{2t}}\) incorrectly, but most were able to earn some further method marks for substituting into their integrated function. The majority of candidates who substituted (\(0\), \(10\)) into their integrated function knew that \({{\text{e}}^0} = 1\) .

Syllabus sections

Topic 6 - Calculus » 6.5 » Anti-differentiation with a boundary condition to determine the constant term.

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