Question

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge, his velocity is horizontal with magnitude $9.0\text{m/s}$. Find the motorcycle’s distance from the edge of the cliff and velocity after $0.5\text{s}$.

• A. Distance $=\frac{349}{4}\text{m}$, Velocity $=\sqrt{106}\text{m/s}$
• B. Distance $=\frac{\sqrt{349}}{2}\text{m}$, Velocity $=\sqrt{106}\text{m/s}$
• C. Distance $=\frac{\sqrt{349}}{4}\text{m}$, Velocity $=106\text{m/s}$
• D. Distance $=\frac{\sqrt{349}}{4}\text{m}$, Velocity $=\sqrt{106}\text{m/s}$

Answer 1

The correct option is D Distance $=\frac{\sqrt{349}}{4}\text{m}$, Velocity $=\sqrt{106}\text{m/s}$



At $t=0.50\text{s}$, the $x$ and $y$-coordinates are
$x=v_{0}t=\left(9.0\text{m/s}\right)\left(0.50\text{s}\right)=4.5\text{m}$
$y=-\frac{1}{2}g{t}^2=-\frac{1}{2}\left(10{\text{m/s}}^2\right)(0.50\text{s}{)}^2=-\frac{5}{4}m$

The negative value of $y$ shows that the motorcycle is below its starting point.

The motorcycle’s distance from the origin at this time $r=\sqrt{x^2+y^2}=\sqrt{{\left(\frac{9}{2}\right)}^2+{\left(\frac{5}{4}\right)}^2}=\frac{\sqrt{349}}{4}\text{m}$.

The components of velocity at this time are
$v_x=v_0=9\text{m/s},v_y=-gt=(-10\text{m/s})\left(0.50\text{s}\right)=-5\text{m/s}$
The speed (magnitude of velocity) at this time is
$v=\sqrt{v_x^2+v_y^2}=\sqrt{\left(9{\text{m/s}}^2\right)+(-5{\text{m/s}}^2)}=\sqrt{106}\text{m/s}$