IB Mathematics SL 5.9 Kinematic problems AA SL Paper 1- Exam Style Questions- New Syllabus
Question
(i) maximum velocity;
(ii) maximum speed.
(ii) Find the distance travelled by the object in the first \( T \) seconds.
Most-appropriate topic codes (Mathematics: analysis and approaches guide):
• SL 5.9: Kinematic problems involving displacement \( s \), velocity \( v \), acceleration \( a \) and total distance travelled — Part a(ii), b, c
• SL 5.5: Introduction to integration as antidifferentiation; link between anti-derivatives, definite integrals and area — Part b(ii), c
▶️ Answer/Explanation
(a)(i) Maximum velocity:
At local maximum \( t = 1 \):
\( v(1) = 30 + 20(1) – 10(1)^2 = 40 \).
Maximum velocity \( = 40\ \text{ms}^{-1} \).
(a)(ii) Maximum speed:
Speed \( = |v(t)| \). Check critical points and boundaries:
\( v(0) = 30 \implies \text{speed } 30 \)
\( v(1) = 40 \implies \text{speed } 40 \)
\( v(5) = 30 + 20(5) – 10(25) = -120 \implies \text{speed } 120 \).
Maximum speed \( = 120\ \text{ms}^{-1} \).
(b)(i) Value of \( T \):
Direction changes when \( v(t) = 0 \). From the graph, this occurs at \( T = 3 \) seconds.
(b)(ii) Distance in first \( T = 3 \) seconds:
Since \( v(t) \ge 0 \) for \( 0 \le t \le 3 \):
\( \text{Distance} = \int_{0}^{3} (30 + 20t – 10t^2) \, dt \)
\( = \left[ 30t + 10t^2 – \frac{10}{3}t^3 \right]_{0}^{3} \)
\( = (90 + 90 – 90) – 0 = 90 \).
Distance travelled \( = 90 \) metres.
(c) Return to initial position:
The object returns to its initial position if total displacement \( = 0 \).
Total displacement \( = \int_{0}^{5} v(t) \, dt \)
\( = \left[ 30t + 10t^2 – \frac{10}{3}t^3 \right]_{0}^{5} \)
\( = (150 + 250 – \frac{1250}{3}) = 400 – 416.67 = -16.67 \).
Since displacement \( \ne 0 \), the object does not return to its initial position.