Vectors & Motion in Two Dimensions (College Board AP® Physics 1: Algebra-Based): Revision Note

A carrier jet is an aircraft that takes off from a large boat called an aircraft carrier. The jet is catapulted from the aircraft carrier to reach the velocity required for takeoff on such a short runway.

At the moment of takeoff, a jet has a horizontal acceleration of $25\mathrm{m}/{\mathrm{s}}^2$ and a vertical acceleration of $11\mathrm{m}/{\mathrm{s}}^2$. The jet's initial velocity is $64\mathrm{m}/\mathrm{s}$ in the horizontal direction and $19\mathrm{m}/\mathrm{s}$ in the vertical direction.

Calculate the resultant final velocity of the jet after $7\mathrm{s}$.

Answer:

Step 1: Analyze the scenario

• To find the resultant final velocity, the component final velocities must first be calculated

Step 2: List the known quantities

• Taking forward and upward to be positive

• Horizontal acceleration, $a_x=25\mathrm{m}/{\mathrm{s}}^2$

• Horizontal initial velocity, $v_{x0}=64\mathrm{m}/\mathrm{s}$

• Vertical acceleration, $a_y=11\mathrm{m}/{\mathrm{s}}^2$

• Vertical initial velocity, $v_{y0}=19\mathrm{m}/\mathrm{s}$

• Time interval, $t=7\mathrm{s}$

Step 3: Select a suitable kinematic equation to calculate the final velocity components

• Displacement is not required

$v_x=v_{x0}+a_{x}t$

$v_x=64+\left(25·7\right)=239\mathrm{m}/\mathrm{s}$

$v_y=19+\left(11·7\right)=96\mathrm{m}/\mathrm{s}$

Step 4: Calculate the magnitude of the resultant final velocity

$v_{x+y}=\sqrt{{239}^2+{96}^2}=258\mathrm{m}/\mathrm{s}$

Step 5: Calculate the direction of the resultant final velocity

$v_{x+y}={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$

$v_{x+y}={\tan }^{-1}\left(\frac{96}{239}\right)=22°$ from the horizontal

Step 6: State the resultant final velocity

$v_{x+y}=258\mathrm{m}/\mathrm{s}\mathrm{at}22°\mathrm{to}\mathrm{the}\mathrm{horizontal}$