Jet Ski

#316 / Dynamics Medium

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 deceleration, what is the total distance traveled by the jet ski before coming to a rest?

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
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, let’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, let’s 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


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