## College Physics (4th Edition)

(a) The distance traveled during this time is $223.5~m$ (b) The acceleration magnitude is $0.99~m/s^2$
(a) $\Delta x = v_{ave}~t$ $\Delta x = \frac{v_f+v_0}{2}~t$ $\Delta x = (\frac{27.3~m/s+17.4~m/s}{2})~(10.0~s)$ $\Delta x = 223.5~m$ The distance traveled during this time is $223.5~m$ (b) We can find the acceleration: $v_f = v_0+at$ $a = \frac{v_f-v_0}{t}$ $a = \frac{27.3~m/s-17.4~m/s}{10.0~s}$ $a = 0.99~m/s^2$ The acceleration magnitude is $0.99~m/s^2$