Solved: Note: In this question, distance is in metres and time is in seconds.A particle P... [IB Mathematics Applications & Interpretation (Math AI)]

Question

SLPaper 2

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 $a = 0$ .

[2]

Verified

Solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$t = \frac{2}{3}{\text{ (exact), }}0.667,{\text{ }}t = 4$ A1A1 N2

[2 marks]

Hence or otherwise, find all possible values of $t$ for which the velocity of P is decreasing.

[2]

Verified

Solution

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]

Find an expression for the velocity of P at time $t$ .

[6]

Verified

Solution

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]

Find the total distance travelled by P when its velocity is increasing.

[4]

Verified

Solution

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]

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 − 14 t + 8, for 0 ⩽ t ⩽ 5.When t = 0, the velocity of P is 3 m s − 1 .

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