A cheetah is hunting. Its prey runs for 2.88 s at a constant velocity of +8.93 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time? (Give the magnitude of the acceleration
Dc = Dp
0.5a*t^2 = 8.93 * 2.88
0.5a*2.88^2 = 25.72
Solve for a.
To find the constant acceleration the cheetah must maintain, we can use the kinematic equation:
\[ d = ut + \frac{1}{2}at^2 \]
where:
- \(d\) is the distance traveled,
- \(u\) is the initial velocity,
- \(t\) is the time taken, and
- \(a\) is the acceleration.
We can first calculate the distance the prey traveled using the formula:
\[ d = ut \]
Substituting the given values, we have:
\[ d = (8.93 \, \text{m/s})(2.88 \, \text{s}) \]
Simplifying, we find:
\[ d = 25.7024 \, \text{m} \]
Since the cheetah starts from rest, the initial velocity (\(u\)) is 0 m/s.
Now, we can set the distance traveled by the cheetah equal to the distance traveled by the prey:
\[ d = ut + \frac{1}{2}at^2 \]
Substituting the known values, we have:
\[ 25.7024 \, \text{m} = 0 + \frac{1}{2}a(2.88 \, \text{s})^2 \]
Simplifying:
\[ 25.7024 \, \text{m} = 0 + 4.1472a \]
Rearranging the equation to solve for \(a\), we get:
\[ a = \frac{25.7024 \, \text{m}}{4.1472 \, \text{s}^2} \]
Evaluating this expression:
\[ a \approx 6.1971 \, \text{m/s}^2 \]
Therefore, the cheetah must maintain a constant acceleration of approximately 6.1971 m/s^2 to run the same distance as its prey in the given time.