(a) A magazine article states that cheetahs are the fastest sprinters in the animal world and that
a cheetah was observed to accelerate from rest to 70 kilometers per hour in 2 seconds. What
average acceleration in meters per second squared does this require?
(b) The article also says the cheetah covered 60 meters during this 2 second interval. How large a
constant acceleration is implied by this statement? Does it agree with your result in (a)?
(c) Accelerations substantially greater than g are difficult for an animal or automobile to attain, because
there is a tendency to slip even on very rough ground with larger acceleration. Given this
information, can you guess which number is wrong in the article?
(a) To find the average acceleration, we can use the formula:
acceleration = change in velocity / time
The change in velocity is given as 70 km/h - 0 km/h = 70 km/h. Converting this to m/s, we have:
70 km/h = 70,000 m/3600 s = 19.44 m/s
The time is given as 2 seconds. Now we can calculate the average acceleration:
acceleration = 19.44 m/s / 2 s = 9.72 m/s²
Therefore, the average acceleration required is 9.72 m/s².
(b) In this part, we need to find the constant acceleration implied by the statement that the cheetah covered 60 meters during the 2-second interval.
We can use the following kinematic equation to solve for the acceleration:
distance = initial velocity × time + (1/2) × acceleration × time²
Plugging in the given values:
60 m = 0 m/s × 2 s + (1/2) × acceleration × (2 s)²
Simplifying:
60 m = 2 s² × acceleration
Now we can solve for the acceleration:
acceleration = 60 m / (2 s²) = 30 m/s²
The constant acceleration implied by the statement is 30 m/s².
This differs from the result in (a), which was 9.72 m/s².
(c) Comparing the two values calculated in (a) and (b), it appears that the number wrong in the article is the statement about the cheetah accelerating from rest to 70 kilometers per hour in 2 seconds. The cheetah would need to have a much larger acceleration to reach that speed in such a short time, more in line with the 30 m/s² calculated in (b).
(a) To find the average acceleration in meters per second squared, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time
In this case, the cheetah starts from rest, so the initial velocity is 0 m/s. The final velocity is given as 70 km/h, which needs to be converted to m/s. We know that 1 km/h is equal to 1000 m/3600 s, so:
Final velocity = 70 km/h * (1000 m/3600 s) = 19.44 m/s
The time is given as 2 seconds.
Average acceleration = (19.44 m/s - 0 m/s) / 2 s = 9.72 m/s^2
Therefore, the average acceleration required is 9.72 meters per second squared.
(b) The distance covered can be calculated using the formula:
Distance = (initial velocity * time) + (0.5 * acceleration * time^2)
In this case, the initial velocity is 0 m/s, the time is 2 seconds, and the distance is given as 60 meters.
60 m = (0 m/s * 2 s) + (0.5 * acceleration * (2 s)^2)
60 m = 2 s^2 * acceleration
acceleration = 60 m / (2 s^2)
acceleration = 30 m/s^2
The implied constant acceleration from the distance covered is 30 meters per second squared.
No, this value does not agree with the result in (a) since they have different magnitudes.
(c) Given that accelerations substantially greater than g (acceleration due to gravity) are difficult for an animal or automobile to attain, we can compare the values obtained in (a) and (b) to make an educated guess.
In (a), the average acceleration is 9.72 m/s^2, which is less than the value for g (approximately 9.8 m/s^2). So, the acceleration seems plausible for a cheetah.
In (b), the constant acceleration is 30 m/s^2, which is significantly greater than g. This value implies that the cheetah accelerated much faster than what is typically achievable, which might indicate an error in the article.
Based on this reasoning, it is likely that the number for the distance covered (60 meters) in the article is wrong.