## College Physics (4th Edition)

(a) The cheetah's acceleration is $12~m/s^2$ (b) The distance traveled during this time is $24~m$ (c) The person's acceleration magnitude is $3.0~m/s^2$ The cheetah's acceleration magnitude is four times greater than the runner's acceleration magnitude.
(a) We can find the cheetah's acceleration: $v_f = v_0+at$ $a = \frac{v_f-v_0}{t}$ $a = \frac{24~m/s-0}{2.0~s}$ $a = 12~m/s^2$ The cheetah's acceleration is $12~m/s^2$ (b) $\Delta x = \frac{1}{2}at^2$ $\Delta x = \frac{1}{2}~(12~m/s^2)(2.0~s)^2$ $\Delta x = 24~m$ The distance traveled during this time is $24~m$ (c) We can find the runner's acceleration: $v_f = v_0+at$ $a = \frac{v_f-v_0}{t}$ $a = \frac{6.0~m/s-0}{2.0~s}$ $a = 3.0~m/s^2$ The person's acceleration magnitude is $3.0~m/s^2$ The cheetah's acceleration magnitude is four times greater than the runner's acceleration magnitude.