A cheetah can accelerate from rest to 25.9 m/s in 1.80 s. Assuming the acceleration is constant over the time interval, what is the magnitude of the acceleration of the cheetah?
What is the distance traveled by the cheetah in these 1.80 s?
a =(Vf - Vo) / t,
a = (25.9 - 0) / 1.8 = 14.39m/s^2.
b. d = Vo*t + 0.5*at^2,
d = 0 + 0.5*14.39*(1.8)^2 = 23.3m.
To find the magnitude of acceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
- Initial velocity (rest) = 0 m/s
- Final velocity = 25.9 m/s
- Time = 1.80 s
Substituting the values into the formula:
acceleration = (25.9 - 0) / 1.80
acceleration = 14.39 m/s^2
Therefore, the magnitude of the acceleration of the cheetah is 14.39 m/s^2.
To find the distance traveled by the cheetah, we can use the formula:
distance = (initial velocity x time) + (0.5 x acceleration x time^2)
Given:
- Initial velocity (rest) = 0 m/s
- Time = 1.80 s
- Acceleration = 14.39 m/s^2
Substituting the values into the formula:
distance = (0 x 1.80) + (0.5 x 14.39 x 1.80^2)
distance = 23.50 m
Therefore, the distance traveled by the cheetah in 1.80 s is 23.50 m.