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 be $\vec{v}_{g}$. The distance between London and Toronto is $d$. Let the time taken for the Toronto - London trip be $t_1$ and for the London - Toronto trip be $t_2$. We want $t_1=t_2$.
On the Toronto - London trip, $\vec{v}_{g1}=\vec{v}_{a1}+\vec{v}_{w}$ and $d = |\vec{v}_{g1}|t_1$. On the London - Toronto trip, $\vec{v}_{g2}=\vec{v}_{a2}+\vec{v}_{w}$ and $d = |\vec{v}_{g2}|t_2$. Since $t_1 = t_2$, we have $|\vec{v}_{g1}|=|\vec{v}_{g2}|$.

Step2: Resolve vectors

If we assume the direction from Toronto to London as the positive $x$ - axis. Let $\vec{v}_{a1}=v_{a1x}\hat{i}+v_{a1y}\hat{j}$ and $\vec{v}_{a2}=v_{a2x}\hat{i}+v_{a2y}\hat{j}$, $\vec{v}_{w}=v_{wx}\hat{i}+v_{wy}\hat{j}$. Then $\vec{v}_{g1}=(v_{a1x} + v_{wx})\hat{i}+(v_{a1y}+v_{wy})\hat{j}$ and $\vec{v}_{g2}=(v_{a2x}+v_{wx})\hat{i}+(v_{a2y}+v_{wy})\hat{j}$. Using $|\vec{v}_{g1}| = |\vec{v}_{g2}|$, we get $(v_{a1x}+v_{wx})^2+(v_{a1y}+v_{wy})^2=(v_{a2x}+v_{wx})^2+(v_{a2y}+v_{wy})^2$. Expanding and simplifying (assuming we know the components of $\vec{v}_{w}$ and $\vec{v}_{a1}$), we can find $\vec{v}_{a2}$.

For the second part:

Step1: Recall acceleration formula

The acceleration $\vec{a}=\frac{\Delta\vec{v}}{\Delta t}$. The plane changes its direction by $90^{\circ}$ while maintaining the same ground - speed $v$. Let the initial velocity be $\vec{v}_1 = v\hat{i}$ and the final velocity be $\vec{v}_2=v\hat{j}$ (assuming appropriate coordinate system).

Step2: Calculate acceleration

$\Delta\vec{v}=\vec{v}_2-\vec{v}_1=v\hat{j}-v\hat{i}$. Given $\Delta t = 7.5\ s$. Then $\vec{a}=\frac{v\hat{j}-v\hat{i}}{7.5}$. The magnitude of the acceleration $a=\frac{\sqrt{v^{2}+(-v)^{2}}}{7.5}=\frac{\sqrt{2}v}{7.5}$.

Answer:

For the first part, the new air - velocity $\vec{v}_{a2}$ can be found by solving the 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 is $\frac{\sqrt{2}v}{7.5}\ m/s^{2}$ (where $v$ is the ground - speed of the plane).