Date | November 2017 | Marks available | 2 | Reference code | 17N.2.sl.TZ0.9 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Write down | 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=3t2−14t+8, for 0⩽.
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]