Answer
(a) The distance traveled during this time is $223.5~m$
(b) The acceleration magnitude is $0.99~m/s^2$
Work Step by Step
(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 $0.99~m/s^2$
