IBDP Maths AA: Topic SL 5.3 Derivative of f(x): IB style Questions SL Paper 2

Detail Solution

(a) Finding the smallest value of 

 when the particle changes direction.

The particle changes direction when its velocity 

, and this change must be confirmed by checking the sign of the acceleration (the derivative of 

) to ensure it is a point where the motion reverses (i.e., a turning point).

First, set the velocity function to zero:

$v=2sin(0.5t) +0.3t-2=0$

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 

. From the graph:
The velocity starts negative (below the 

-axis) at 

.
 It crosses the 

-axis (where 

) and reaches a peak, then crosses again before descending.
 The first crossing appears to occur between 

 and 

, and the second crossing is around 

 to 

.

To find the exact value, solve 

. Let’s approximate by testing values near the first crossing:
 At 

: 

 (positive).
 At 

: 

 (negative).

Since 

 changes from negative to positive between 

 and 

, the root lies in this interval. Let’s refine further:
 At 

: 

 (negative).
 At 

: 

 (positive).

The velocity changes from negative to positive between 

 and 

, so the root is approximately 

. To confirm this is a direction change, compute the acceleration 

:

$$a=\frac{d}{dt}(2\sin (0.5t)+0.3t–2)=2\cdot 0.5\cos (0.5t)+0.3=\cos (0.5t)+0.3$$

At 

:

(positive).

Since 

, 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 

 when the particle changes direction is approximately 

 seconds.

(b) Finding the range of values of 

 for which the displacement of the particle is increasing.

The displacement 

 is increasing when the velocity 

. From the graph:

 from 

 to the first zero (around 

). 

 from 

 to the second zero (around 

). 

 from 

 to 

.

The displacement 

 is the integral of the velocity function:

Compute the indefinite integral:

The particle passes through 

 at 

, and assuming 

 is the origin (displacement 

 at 

):

So, the displacement function is:

Now, evaluate at 

:

$$s(10)=-4\cos (0.5\cdot 10)+0.15\cdot {10}^2–2\cdot 10+4$$

$$=-4\cos (5)+0.15\cdot 100–20+4$$

$$\cos (5)\approx 0.2837$$

$$s(10)=-4\cdot 0.2837+15–20+4$$

$$=-1.1348–1$$

$$\approx -2.1348$$

The displacement at 

 is approximately 

 meters (relative to 

, negative indicating displacement to the left of 

.

————Markscheme—————–

(a) recognition that velocity is zero 
$v =  2sin (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