IBDP Maths AA: Topic SL 5.9: Kinematic problems: IB style Questions HL Paper 2

Detailed Solution

(a) Finding the smallest value of \( t \) when the particle changes direction.

The particle changes direction when its velocity \( v = 0 \), and this change must be confirmed by checking the sign of the acceleration (the derivative of \( v \)) to ensure it is a point where the motion reverses (i.e., a turning point).

First, set the velocity function to zero:
\[ v = 2\sin(0.5t) + 0.3t – 2 = 0 \]

This equation is transcendental and may require numerical methods or graphical analysis to solve exactly. However, we can use the graph provided to identify where \( v = 0 \). From the graph:
The velocity starts negative (below the \( t \)-axis) at \( t = 0 \).
 It crosses the \( t \)-axis (where \( v = 0 \)) and reaches a peak, then crosses again before descending.
 The first crossing appears to occur between \( t = 2 \) and \( t = 3 \), and the second crossing is around \( t = 7 \) to \( t = 8 \).

To find the exact value, solve \( 2\sin(0.5t) + 0.3t – 2 = 0 \). Let’s approximate by testing values near the first crossing:
 At \( t = 2 \): \( v = 2\sin(0.5 \cdot 2) + 0.3 \cdot 2 – 2 = 2\sin(1) + 0.6 – 2 \approx 2 \cdot 0.8415 + 0.6 – 2 \approx 1.683 + 0.6 – 2 = 0.283 \) (positive).
 At \( t = 1 \): \( v = 2\sin(0.5 \cdot 1) + 0.3 \cdot 1 – 2 = 2\sin(0.5) + 0.3 – 2 \approx 2 \cdot 0.4794 + 0.3 – 2 \approx 0.9588 + 0.3 – 2 = -0.7412 \) (negative).

Since \( v \) changes from negative to positive between \( t = 1 \) and \( t = 2 \), the root lies in this interval. Let’s refine further:
 At \( t = 1.5 \): \( v = 2\sin(0.5 \cdot 1.5) + 0.3 \cdot 1.5 – 2 = 2\sin(0.75) + 0.45 – 2 \approx 2 \cdot 0.6816 + 0.45 – 2 \approx 1.3632 + 0.45 – 2 = -0.1868 \) (negative).
 At \( t = 1.7 \): \( v = 2\sin(0.5 \cdot 1.7) + 0.3 \cdot 1.7 – 2 = 2\sin(0.85) + 0.51 – 2 \approx 2 \cdot 0.7506 + 0.51 – 2 \approx 1.5012 + 0.51 – 2 = 0.0112 \) (positive).

The velocity changes from negative to positive between \( t = 1.5 \) and \( t = 1.7 \), so the root is approximately \( t \approx 1.6 \). To confirm this is a direction change, compute the acceleration \( a = \frac{dv}{dt} \):
\[ a = \frac{d}{dt} (2\sin(0.5t) + 0.3t – 2) = 2 \cdot 0.5 \cos(0.5t) + 0.3 = \cos(0.5t) + 0.3 \]

At \( t = 1.6 \):
\[ a = \cos(0.5 \cdot 1.6) + 0.3 = \cos(0.8) + 0.3 \approx 0.6967 + 0.3 = 0.9967 \] (positive).

Since \( a > 0 \), the velocity is increasing through zero, indicating a change from negative to positive velocity (the particle was moving left, stopped, and started moving right). This is a direction change. The graph also suggests this is the first such point. Thus, the smallest \( t \) when the particle changes direction is approximately \( t \approx 1.6 \) seconds.

(b) Finding the range of values of \( t \) for which the displacement of the particle is increasing.

The displacement \( s(t) \) is increasing when the velocity \( v > 0 \). From the graph:
\( v < 0 \) from \( t = 0 \) to the first zero (around \( t \approx 1.6 \)).
 \( v > 0 \) from \( t \approx 1.6 \) to the second zero (around \( t \approx 7.5 \)).
 \( v < 0 \) from \( t \approx 7.5 \) to \( t = 10 \).

To find the exact second zero, solve \( 2\sin(0.5t) + 0.3t – 2 = 0 \) again:
 At \( t = 7 \): \( v = 2\sin(0.5 \cdot 7) + 0.3 \cdot 7 – 2 = 2\sin(3.5) + 2.1 – 2 \approx 2 \cdot (-0.3508) + 2.1 – 2 \approx -0.7016 + 0.1 = -0.6016 \) (negative).
 At \( t = 8 \): \( v = 2\sin(0.5 \cdot 8) + 0.3 \cdot 8 – 2 = 2\sin(4) + 2.4 – 2 \approx 2 \cdot (-0.7568) + 2.4 – 2 \approx -1.5136 + 0.4 = -1.1136 \) (negative).
 At \( t = 7.5 \): \( v = 2\sin(0.5 \cdot 7.5) + 0.3 \cdot 7.5 – 2 = 2\sin(3.75) + 2.25 – 2 \approx 2 \cdot (0.065) + 2.25 – 2 \approx 0.1396 + 0.25 = 0.38 \) (positive).
 At \( t = 7.4 \): \( v = 2\sin(0.5 \cdot 7.4) + 0.3 \cdot 7.4 – 2 = 2\sin(3.7) + 2.22 – 2 \approx 2 \cdot (-0.1577) + 2.22 – 2 \approx -0.3154 + 0.22 = -0.0954 \) (negative).

The second zero is between \( t = 7.4 \) and \( t = 7.5 \), approximately \( t \approx 7.45 \). Thus, \( v > 0 \) from \( t \approx 1.6 \) to \( t \approx 7.45 \). The range of \( t \) for which the displacement is increasing is approximately \( 1.6 < t < 7.45 \).

(c) Finding the displacement of the particle relative to \( O \) when \( t = 10 \).

The displacement \( s(t) \) is the integral of the velocity function:
\[ s(t) = \int v \, dt = \int (2\sin(0.5t) + 0.3t – 2) \, dt \]

Compute the indefinite integral:
\[ \int 2\sin(0.5t) \, dt = 2 \cdot \frac{-\cos(0.5t)}{0.5} = -4\cos(0.5t) \]
\[ \int 0.3t \, dt = 0.3 \cdot \frac{t^2}{2} = 0.15t^2 \]
\[ \int -2 \, dt = -2t \]
\[ s(t) = -4\cos(0.5t) + 0.15t^2 – 2t + C \]

The particle passes through \( O \) at \( t = 0 \), and assuming \( O \) is the origin (displacement \( s = 0 \) at \( t = 0 \)):
\[ s(0) = -4\cos(0) + 0.15 \cdot 0^2 – 2 \cdot 0 + C = -4 \cdot 1 + C = 0 \]
\[ C = 4 \]

So, the displacement function is:
\[ s(t) = -4\cos(0.5t) + 0.15t^2 – 2t + 4 \]

Now, evaluate at \( t = 10 \):
\[ s(10) = -4\cos(0.5 \cdot 10) + 0.15 \cdot 10^2 – 2 \cdot 10 + 4 \]
\[ = -4\cos(5) + 0.15 \cdot 100 – 20 + 4 \]
\[ \cos(5) \approx 0.2837 \]
\[ s(10) = -4 \cdot 0.2837 + 15 – 20 + 4 \]
\[ = -1.1348 – 1 \]
\[ \approx -2.1348 \]

The displacement at \( t = 10 \) is approximately \(-2.13\) meters (relative to \( O \), negative indicating displacement to the left of \( O \).

…………………….Markscheme…………………….

Solution: –

(a) recognition that velocity is zero
$v = 2\sin(0.5t) + 0.3t – 2 = 0$
$t = 1.68694…$
$t = 1.69$

(b) recognition that $v > 0$
$1.68694… < t < 6.11857…$
$1.69 < t < 6.12$

(c) attempt to substitute into the total displacement formula (condone missing or incorrect limits, and absence of dt)
$\int_{0}^{10} (2\sin(0.5t) + 0.3t – 2) dt$ OR $\int_{0}^{10} v(t) dt$
= -2.13464…
= -2.13 (m)