cheetahs, the fastest of the great cats, can reach 45 mph in 2.0 s starting from rest. assuming that they have constant acceleration of 33 ft/s throughout that time, find the distance (in ft) they travel during that time.
dude i think they were wanting the answer but ur equation doesnt go with any of the answers that does go with the question ive done the work.
To find the distance traveled by the cheetahs during that time, we can use the equation:
\[ d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]
where:
- \(d\) is the distance traveled,
- \(v_i\) is the initial velocity,
- \(t\) is the time, and
- \(a\) is the acceleration.
Given:
\(v_i = 0\) (starting from rest)
\(t = 2.0 \, \text{s}\)
\(a = 33 \, \text{ft/s}\)
Substituting these values into the equation, we get:
\[ d = 0 \cdot 2.0 + \frac{1}{2} \cdot 33 \cdot (2.0)^2 \]
Simplifying this expression, we get:
\[ d = \frac{1}{2} \cdot 33 \cdot 4.0 \]
\[ d = 66 \, \text{ft} \]
Therefore, the cheetahs travel a distance of 66 feet during that time.
To find the distance traveled by the cheetah, we can use the equation:
distance = initial velocity × time + 0.5 × acceleration × time²
First, let's convert the initial speed of the cheetah from mph to ft/s:
45 mph = 45 × 1.47 ft/s (1 mph = 1.47 ft/s)
So, the initial velocity (v) of the cheetah is 45 × 1.47 = 66.15 ft/s.
Now, let's substitute the given values into the equation:
distance = 66.15 ft/s × 2.0 s + 0.5 × 33 ft/s² × (2.0 s)²
distance = 132.3 ft + 0.5 × 33 ft/s² × 4.0 s²
Since 0.5 × 33 ft/s² = 16.5 ft/s², we can simplify further:
distance = 132.3 ft + 16.5 ft/s² × 4.0 s²
distance = 132.3 ft + 16.5 ft/s² × 16.0 s²
Calculating further,
distance = 132.3 ft + 264 ft
distance = 396.3 ft
Therefore, during that time, the cheetah travels a distance of 396.3 ft.
s = 1/2 at^2
now plug in your acceleration and time...