IB Mathematics AHL 5.13 Kinematic problems involving displacement-AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
Set \( v = -2t^2 + 16t – 24 = 0 \).
Divide by \(-2\): \( t^2 – 8t + 12 = 0 \).
Factorize: \( (t – 2)(t – 6) = 0 \).
So, \( t = 2 \) or \( t = 6 \) seconds.
Explanation:
Instantaneous rest occurs when velocity is zero. Solve the quadratic equation.
Result:
\( t = 2 \) seconds, \( t = 6 \) seconds

(b)
Acceleration \( a = \frac{dv}{dt} \).
Differentiate \( v = -2t^2 + 16t – 24 \): \( a = -4t + 16 \).
At \( t = 6 \): \( a = -4 \times 6 + 16 = -24 + 16 = -8 \) m s\(^{-2}\).
Magnitude = \( |-8| = 8 \) m s\(^{-2}\).
Explanation:
Acceleration is the derivative of velocity; magnitude is the absolute value.
Result:
\( 8 \) m s\(^{-2}\)

(c)
Speed is \( |v| \). Find critical points: \( \frac{dv}{dt} = -4t + 16 = 0 \) \( \Rightarrow t = 4 \).
Evaluate \( v \) at \( t = 0 \), \( t = 4 \), \( t = 6 \):
\( t = 0 \): \( v = -24 \), speed = \( 24 \) m s\(^{-1}\);
\( t = 4 \): \( v = -2 \times 16 + 16 \times 4 – 24 = 8 \) m s\(^{-1}\);
\( t = 6 \): \( v = 0 \).
Greatest speed is \( 24 \) m s\(^{-1}\) at \( t = 0 \).
Explanation:
Maximum speed is the highest \( |v| \) in the interval.
Result:
\( 24 \) m s\(^{-1}\)

(d)
Displacement \( x = \int v \, dt \).
Integrate \( v = -2t^2 + 16t – 24 \):
\( x = -\frac{2t^3}{3} + 8t^2 – 24t + c \).
At \( t = 0 \), \( x = 0 \): \( c = 0 \).
So, \( x = -\frac{2t^3}{3} + 8t^2 – 24t \).
Explanation:
Displacement is the integral of velocity with initial condition at origin.
Result:
\( x = -\frac{2t^3}{3} + 8t^2 – 24t \)

(e)
Total distance is \( \int_0^4 |v| \, dt \).
Since \( v = -2t^2 + 16t – 24 \) has zeros at \( t = 2, 6 \), on \( 0 \leq t \leq 4 \), \( v < 0 \) on \( [0, 2] \) and \( v > 0 \) on \( [2, 4] \). Hence:
\( \int_0^4 |v| \, dt = \int_0^2 -(v) \, dt + \int_2^4 v \, dt \).
First part:
\( \int_0^2 (2t^2 – 16t + 24) \, dt = [ \frac{2t^3}{3} – 8t^2 + 24t ]_0^2 = \frac{2 \times 8}{3} – 8 \times 4 + 24 \times 2 = \frac{16}{3} \).
Second part:
\( \int_2^4 (-2t^2 + 16t – 24) \, dt = [ -\frac{2t^3}{3} + 8t^2 – 24t ]_2^4 = (-\frac{128}{3} + 128 – 96) – (-\frac{16}{3} + 32 – 48) = \frac{32}{3} \).
Therefore, total distance = \( \frac{64}{3} + \frac{32}{3} = \frac{96}{3} = 32 \) m.
Explanation:
Split the integral of \( |v| \) at the sign change \( t = 2 \); add the magnitudes of the two signed displacements.
Result:
\( 32 \) m