IB DP Math MAA SL : IB Style Mock Exams – Set 1 Paper 2
Question
A particle moves along a straight line such that its velocity, \( v \), ms-1, is given by \( v(t) = 5t e^{-1.2t} \), for \( t > 0 \).
(a) On the grid below, sketch the graph of \( v \), for \( 0 \leq t \leq 3 \).
(b) Find the distance travelled by the particle in the first 3 seconds.
(c) Find the maximum velocity of the particle in the first 3 seconds.
▶️ Answer/Explanation
Part (a): Sketch of the graph
Key features:
- \( v(0) = 0 \) → graph starts at origin
- Function increases initially, reaches a maximum, then decreases
- Always positive for \( t > 0 \)
- Approaches 0 as \( t \) increases
The sketch is shown below:
Part (b): Distance travelled
Step 1: Use integral of velocity
Since velocity is positive, distance = displacement:
\( d = \int_{0}^{3} 5t e^{-1.2t} \, dt \)
Step 2: Evaluate
Using a calculator:
\( d \approx 3.04 \text{ m} \)
Answer (b):
\( \boxed{3.04 \text{ m}} \)
Part (c): Maximum velocity
Step 1: Differentiate velocity
Maximum occurs when \( \frac{dv}{dt} = 0 \)
Step 2: Solve (using calculator)
\( t \approx 0.833 \)
Step 3: Substitute into velocity
\( v(0.833) \approx 1.53 \)
Answer (c):
\( \boxed{1.53 \text{ ms}^{-1}} \)