Aircraft Carrier

#322 / Dynamics Easy

Imagine fighter jets on an aircraft carrier are launched off a 300 m runway from rest. If the jets’ acceleration is 10 m/s^2, what is the minimum takeoff speed?

Expand Hint
$$$v^2=v_0^2+2a_0(s-s_0)$$$
where $$v$$ is the velocity along the direction of travel, $$v_0$$ is the velocity at time $$t_0$$ , $$a_0$$ is constant acceleration, $$s$$ is the displacement at time $$t$$ along the line of travel, and $$s_0$$ is the displacement at time $$t_0$$ .
Hint 2
Solve for $$v$$ .
For constant acceleration, the equation for velocity as a function of position:
$$$v^2=v_0^2+2a_0(s-s_0)$$$
where $$v$$ is the velocity along the direction of travel, $$v_0$$ is the velocity at time $$t_0$$ , $$a_0$$ is constant acceleration, $$s$$ is the displacement at time $$t$$ along the line of travel, and $$s_0$$ is the displacement at time $$t_0$$ . Since the jets takeoff from rest, $$v_0=0$$ :
$$$v^2=(0m/s)^2+2(10m/s^2)(300m)$$$
$$$v=\sqrt{2(10m/s^2)(300m)}=\sqrt{6,000m^2/s^2}=77.46\:m/s$$$
77.46 m/s


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