## Jet Ski

A jet ski rider successfully stops their vessel from crashing into some rocks by reducing the velocity uniformly from 15 m/s to 5 m/s during the first 10 m. Assuming constant deacceleration, what is the total distance traveled by the jet ski before coming to a rest?

##
__
__**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**

Use the above equation to first solve for acceleration, then again for the total distance.

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
$$t_0$$
. First, letâ€™s solve for acceleration from the first 10 m scenario:

$$$(5m/s)^2=(15m/s)^2+2a(10m)$$$

$$$a=\frac{(5m/s)^2-(15m/s)^2}{2(10m)}=\frac{(25-225)m^2/s^2}{20m}=\frac{-200m}{20s^2}=-10\:m/s^2$$$

Since acceleration is constant, we can reuse the same equation to analyze the whole picture:

$$$(0m/s)^2=(15m/s)^2+2(-10m/s^2)(x)$$$

where
$$x$$
is the total distance traveled. Therefore,

$$$x=\frac{-(15m/s)^2}{2(-10m/s^2)}=\frac{-225m^2/s^2}{-20m/s^2}=11.25\:m$$$

11.25 m