Question

kinematics problem set 2 s2 2025
/20
name: rihanna duldul
on the return flight from london to toronto the pilot must adjust the air velocity so that the flight back takes the same time as the toronto - london trip. assuming that the wind has remained the same, find the new air velocity of the airplane. use appropriate variables. the solution is neat and clear. 6 marks
as soon as the plane is approaching toronto, the pilot has to modify the flight path, diverting toward hamilton airport due to a snow - storm. the pilot maintains the same ground speed changing direction by 90.0 degrees. if the maneuver took 7.5 seconds, find the acceleration of the plane during this time. use appropriate variables. the solution is neat and clear. 3 marks

Explanation:

Step1: Define variables

Let the air - velocity of the airplane be $\vec{v}_{a}$, the wind - velocity be $\vec{v}_{w}$, the ground - velocity on the London - Toronto trip be $\vec{v}_{g1}$ and on the return trip be $\vec{v}_{g2}$. Let the distance between London and Toronto be $d$. The time taken for the London - Toronto trip is $t_1=\frac{d}{\vert\vec{v}_{g1}\vert}$ and for the return trip is $t_2=\frac{d}{\vert\vec{v}_{g2}\vert}$. We know $t_1 = t_2$. Also, $\vec{v}_{g1}=\vec{v}_{a1}+\vec{v}_{w}$ and $\vec{v}_{g2}=\vec{v}_{a2}+\vec{v}_{w}$ (vector addition). Since $t_1 = t_2$ and $d$ is constant, $\vert\vec{v}_{g1}\vert=\vert\vec{v}_{g2}\vert$.

Step2: Analyze vector components

If we assume the direction from London to Toronto as the positive $x$ - axis (for simplicity of vector analysis), and represent the vectors in component form. Let $\vec{v}_{a1}=v_{a1x}\hat{i}+v_{a1y}\hat{j}$, $\vec{v}_{a2}=v_{a2x}\hat{i}+v_{a2y}\hat{j}$, and $\vec{v}_{w}=v_{wx}\hat{i}+v_{wy}\hat{j}$. Then $\vec{v}_{g1}=(v_{a1x} + v_{wx})\hat{i}+(v_{a1y}+v_{wy})\hat{j}$ and $\vec{v}_{g2}=(v_{a2x}+v_{wx})\hat{i}+(v_{a2y}+v_{wy})\hat{j}$. Since $\vert\vec{v}_{g1}\vert=\vert\vec{v}_{g2}\vert$, $(v_{a1x} + v_{wx})^2+(v_{a1y}+v_{wy})^2=(v_{a2x}+v_{wx})^2+(v_{a2y}+v_{wy})^2$. Expanding and simplifying using the fact that the wind velocity is constant, we can find the relationship between the components of $\vec{v}_{a1}$ and $\vec{v}_{a2}$.

For the second part:

Step3: Define acceleration formula

Acceleration $\vec{a}=\frac{\Delta\vec{v}}{\Delta t}$. The initial velocity $\vec{v}_i$ and the final velocity $\vec{v}_f$ are related to the ground - speed (which is constant, say $v$). If the initial direction is along one axis (say the $x$ - axis) and the final direction is perpendicular (say the $y$ - axis), $\vec{v}_i = v\hat{i}$ and $\vec{v}_f=v\hat{j}$. Then $\Delta\vec{v}=\vec{v}_f-\vec{v}_i=v\hat{j}-v\hat{i}$.

Step4: Calculate acceleration

Given $\Delta t = 7.5\ s$, $\vec{a}=\frac{v\hat{j}-v\hat{i}}{7.5}$. The magnitude of the acceleration $a=\frac{\sqrt{v^{2}+(-v)^{2}}}{7.5}=\frac{\sqrt{2}v}{7.5}$. But we need to express it in terms of variables. If we assume the ground - speed is $v_g$, then the acceleration magnitude $a=\frac{\sqrt{2}v_g}{7.5}$.

Answer:

For the first part, the new air - velocity $\vec{v}_{a2}$ can be found by solving the vector - equality equation $(v_{a1x} + v_{wx})^2+(v_{a1y}+v_{wy})^2=(v_{a2x}+v_{wx})^2+(v_{a2y}+v_{wy})^2$ for the components of $\vec{v}_{a2}$. For the second part, the magnitude of the acceleration of the plane is $a = \frac{\sqrt{2}v_g}{7.5}$, where $v_g$ is the ground - speed of the plane.