Question

A motorcycle shunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude $9.0m/s$. Find the motorcycle's velocity after $0.5s$.

• A. $\sqrt{106}$ m/s
• B. $106m/s$
• C. $\sqrt{140}$ m/s
• D. $140m/s$

Answer 1

The correct option is A $\sqrt{106}$ m/s

At $t$ $=$ $0.50s$, the $x$ and $y$ - coordinates are
$x=v_{0}t=\left(9.0m/s\right(0.50s)=4.5m$
$y=-\frac{1}{2}g{t}^2=-\frac{1}{2}\left(10m/s^2\right)(0.50s{)}^2=-\frac{5}{4}m$
The negative value of y shows that this time the motorcycle is below its starting point.
The motorcycle's distance from the origin at this time
$r=\sqrt{x^2+y^2}=\sqrt{{\left(\frac{9}{2}\right)}^2+{\left(\frac{5}{4}\right)}^2}=\frac{\sqrt{349}}{4}$ $m$
The components of velocity at this time are.
$v_x=v_0=9.0m/s$
$v_y=-gt=(-10m/s^2)\left(0.50s\right)=-5m/s$
The speed (magnitude of the velocity) at this time is
$v=\sqrt{v_x^2+v_y^2}=\sqrt{(9.0m/s{)}^2+(-5m/s{)}^2}=\sqrt{106}$$m/s$