91Ó°ÊÓ

We can use the formula $$a = \frac{v - u}{t}$$ where \(a\) is acceleration, \(v\) is final velocity, \(u\) is initial velocity, and \(t\) is time. For the cheetah, the initial velocity is 0 (rest), the final velocity is 24 m/s, and the time interval is 2.0 s. So, we have: $$a = \frac{24-0}{2.0}$$ $$a = 12 \,\text{m/s}^2$$ The magnitude of the acceleration of the cheetah is 12 m/s².

One of the kinematic equations is: $$s = ut + \frac{1}{2}at^2$$ where \(s\) is distance, \(u\) is initial velocity, \(t\) is time, and \(a\) is acceleration. For the cheetah, initial velocity is 0, time is 2.0 s, and acceleration is found to be 12 m/s². So, we have: $$s = 0(2.0) + \frac{1}{2}(12)(2.0)^2$$ $$s = 24 \,\text{m}$$ The distance traveled by the cheetah in 2.0 s is 24 m.

Using the same acceleration formula as before: $$a' = \frac{v' - u'}{t'}$$ Where \(a'\), \(v'\), \(u'\) and \(t'\) are the acceleration, final velocity, initial velocity, and time, respectively, for the runner. The initial velocity is 0 (rest), the final velocity is 6.0 m/s, and the time interval is 2.0 s. So we have: $$a' = \frac{6.0 - 0}{2.0}$$ $$a' = 3 \,\text{m/s}^2$$ The magnitude of the acceleration of the runner is 3 m/s². To find the factor by which the cheetah's average acceleration magnitude is greater than that of the runner, divide the magnitude of the cheetah's acceleration by the magnitude of the runner's acceleration: $$ \text{Factor} = \frac{a}{a'} = \frac{12}{3} = 4 $$ The cheetah's average acceleration magnitude is 4 times greater than that of the runner.