IBDP Maths SL 2.9 : Logarithmic functions and their graphs AA HL Paper 2- Exam Style Questions- New Syllabus

(a)(i)

At \(t=0\),

\( v_{\text{air}}=60e^{-0.1(0)}=60e^0=60. \)

Therefore, the airplane lands with a speed of

✅ \(60\text{ m s}^{-1}\)

(a)(ii)

At \(t=0\),

\( v_{\text{car}}=5(0)=0. \)

The car starts from rest.

✅ \(0\text{ m s}^{-1}\)

(b)(i)

Equal speeds occur when

\( 60e^{-0.1t}=5t. \)

Solving this equation (using a calculator or GDC) gives

\( t\approx6.36\text{ s}. \)

✅ \(t=6.36\text{ s}\)

(b)(ii)

Substitute \(t=6.36\) into either velocity function:

\( v=5(6.35564)\approx31.78. \)

Hence, the common speed is

✅ \(31.8\text{ m s}^{-1}\)

(c)

The displacement of the airplane is

\( \int60e^{-0.1t}\,dt=-600e^{-0.1t}+C. \)

The displacement of the car is

\( \int5t\,dt=\frac52t^2+C. \)

Since the initial distance is \(100\) m,

\( d(t)=100+\text{(airplane displacement)}-\text{(car displacement)}. \)

Applying the condition \(d(0)=100\) gives the constant, leading to

\( \boxed{d(t)=-600e^{-0.1t}-\frac52t^2+700.} \)

This function models the separation between the car and the airplane at any time \(t\).

✅ \(d(t)=-600e^{-0.1t}-\dfrac52t^2+700\)

(d)

The car reaches the airplane when

\( d(t)=0. \)

Solve

\( -600e^{-0.1t}-\frac52t^2+700=0. \)

Using a graphical or numerical solver,

\( t\approx15.06\text{ s}. \)

Hence,

✅ \(t=15.1\text{ s}\)

(e)

The distance travelled by the car equals its displacement:

\( \int_0^{15.0586}5t\,dt =\left[\frac52t^2\right]_0^{15.0586}. \)

\( =\frac52(15.0586)^2 \approx566.9. \)

Therefore, the car travels approximately

✅ \(567\text{ m}\)

This is greater than the initial \(100\) m separation because the airplane continues moving forward while slowing down.