## Aircraft Carrier

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

##
__
__**Hint**

**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**

**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