Date | November 2017 | Marks available | 6 | Reference code | 17N.2.sl.TZ0.9 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
Note: In this question, distance is in metres and time is in seconds.
A particle P moves in a straight line for five seconds. Its acceleration at time \(t\) is given by \(a = 3{t^2} - 14t + 8\), for \(0 \leqslant t \leqslant 5\).
When \(t = 0\), the velocity of P is \(3{\text{ m}}\,{{\text{s}}^{ - 1}}\).
Write down the values of \(t\) when \(a = 0\).
Hence or otherwise, find all possible values of \(t\) for which the velocity of P is decreasing.
Find an expression for the velocity of P at time \(t\).
Find the total distance travelled by P when its velocity is increasing.
Markscheme
\(t = \frac{2}{3}{\text{ (exact), }}0.667,{\text{ }}t = 4\) A1A1 N2
[2 marks]
recognizing that \(v\) is decreasing when \(a\) is negative (M1)
eg\(\,\,\,\,\,\)\(a < 0,{\text{ }}3{t^2} - 14t + 8 \leqslant 0\), sketch of \(a\)
correct interval A1 N2
eg\(\,\,\,\,\,\)\(\frac{2}{3} < t < 4\)
[2 marks]
valid approach (do not accept a definite integral) (M1)
eg\(\,\,\,\,\,\)\(v\int a \)
correct integration (accept missing \(c\)) (A1)(A1)(A1)
\({t^3} - 7{t^2} + 8t + c\)
substituting \(t = 0,{\text{ }}v = 3\) , (must have \(c\)) (M1)
eg\(\,\,\,\,\,\)\(3 = {0^3} - 7({0^2}) + 8(0) + c,{\text{ }}c = 3\)
\(v = {t^3} - 7{t^2} + 8t + 3\) A1 N6
[6 marks]
recognizing that \(v\) increases outside the interval found in part (b) (M1)
eg\(\,\,\,\,\,\)\(0 < t < \frac{2}{3},{\text{ }}4 < t < 5\), diagram
one correct substitution into distance formula (A1)
eg\(\,\,\,\,\,\)\(\int_0^{\frac{2}{3}} {\left| v \right|,{\text{ }}\int_4^5 {\left| v \right|} ,{\text{ }}\int_{\frac{2}{3}}^4 {\left| v \right|} ,{\text{ }}\int_0^5 {\left| v \right|} } \)
one correct pair (A1)
eg\(\,\,\,\,\,\)3.13580 and 11.0833, 20.9906 and 35.2097
14.2191 A1 N2
\(d = 14.2{\text{ (m)}}\)
[4 marks]