## Fundamentals of Physics Extended (10th Edition)

The cab has a maximum speed of 305m/min and it starts at rest (0 m/min). Since the acceleration is given in $m/s^2$ (1.22 $m/s^2$), we should convert this speed to m/s. 305m/min * (1min/60seconds) = 5.08m/s The questions asks how far the cab moves while accelerating to full speed so we use: $v^2 = v0^2 + 2a\Delta x$ where v0 = 0, v = 5.08m/s and a = 1.22 $m/s^2)$ $5.08 ^ 2 = 0 ^2 + 2(1.22)\Delta x$ And find that $\Delta x$ is 10.58 meters