A cheetah can accelerate from rest to 40.3 m/s in 2.00 s. Assume that the acceleration is constant over the time interval. A runner
The correct answer and explanation is:
To calculate the acceleration of the cheetah, we can use the following kinematic equation: v=u+atv = u + at
Where:
- vv is the final velocity (40.3 m/s)
- uu is the initial velocity (0 m/s, as the cheetah starts from rest)
- aa is the acceleration (what we need to find)
- tt is the time interval (2.00 s)
Substitute the known values into the equation: 40.3=0+a×2.0040.3 = 0 + a \times 2.00
Simplify: 40.3=2a40.3 = 2a
Solve for aa: a=40.32.00=20.15 m/s2a = \frac{40.3}{2.00} = 20.15 \, \text{m/s}^2
So, the cheetah’s acceleration is 20.15 m/s220.15 \, \text{m/s}^2.
Explanation
The problem involves a cheetah accelerating from rest to a final speed over a given time. The equation used, v=u+atv = u + at, is a basic kinematic equation that describes the motion of an object under constant acceleration. The equation relates the final velocity, initial velocity, acceleration, and time.
Here, the initial velocity uu is zero because the cheetah starts from rest. The final velocity vv is provided as 40.3 m/s, and the time interval tt is given as 2.00 s. The acceleration is constant during this time, and that allows us to use the equation to solve for the acceleration aa.
The formula rearranges to isolate aa, and by substituting the known values, we find that the acceleration is 20.15 m/s220.15 \, \text{m/s}^2. This result means that every second, the cheetah increases its speed by 20.15 meters per second. This is an extremely high acceleration compared to typical animals, highlighting the cheetah’s exceptional ability to accelerate quickly.