Question

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A cheetah can accelerate from rest to \(24 \mathrm{m} / \mathrm{s}\) in $2.0 \mathrm{s}$ Assuming the acceleration is constant over the time interval, (a) what is the magnitude of the acceleration of the cheetah? (b) What is the distance traveled by the cheetah in these \(2.0 \mathrm{s} ?\) (c) A runner can accelerate from rest to \(6.0 \mathrm{m} / \mathrm{s}\) in the same time, $2.0 \mathrm{s} .$ What is the magnitude of the acceleration of the runner? By what factor is the cheetah's average acceleration magnitude greater than that of the runner?

Short answer

Answer: The cheetah's acceleration is 4 times greater than that of the runner.

Step-by-step solution

(a) Calculate the cheetah's acceleration magnitude.

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.

(b) Calculate the distance traveled by the cheetah.

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.

(c) Calculate the runner's acceleration and cheetah's acceleration magnitude compared to the runner's.

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.