Answer
(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.
Work Step by Step
(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.
