The cheetah can reach a top speed of 114 km/h (71 mi/h). While chasing its prey in a short sprint, a cheetah starts from rest and runs 46 m in a straight line, reaching a final speed of 83 km/h.
(a) Determine the cheetah's average acceleration during the short sprint.
(b) Find its displacement at t = 3.0 s. (Assume the cheetah maintains a constant acceleration throughout the sprint.)
Vfinal = 83 km/h = 23.06 m/s
(a) Vfinal^2 = 2*a*X, so
a = Vfinal^2/(2*X)
a = 5.78 m/s^2
(b)At t = 3.0
V = a*t = 17.33 m/s
x = (1/2)a*t^2 = 26.0 m
800×30
To solve this problem, we'll need to use the equations of motion. The equations we'll use are:
v = u + at (Equation 1)
s = ut + (1/2)at^2 (Equation 2)
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
(a) To find the average acceleration during the short sprint, we'll use Equation 1. Since we know the initial velocity (u) is 0 km/h and the final velocity (v) is 83 km/h, we can rearrange the equation to solve for acceleration (a).
v = u + at
83 = 0 + a * t
a = 83 / t
Now we need to find the time it takes for the cheetah to reach its final speed. We are given that the cheetah runs 46 m in a straight line, but we need to convert that to km.
46 m = 0.046 km
Now we can use the equation of motion (Equation 2) to find the time it takes for the cheetah to reach its final speed:
s = ut + (1/2)at^2
0.046 = 0*t + (1/2)a*t^2
0.046 = (1/2)a*t^2
Simplifying, we get:
a * t^2 = 0.092
t^2 = 0.092 / a
Taking the square root of both sides, we get:
t = sqrt(0.092 / a)
Now we can substitute this expression for t into the equation a = 83 / t to solve for the average acceleration:
a = 83 / sqrt(0.092 / a)
Squaring both sides, we get:
a^2 = (83^2 * a^2) / 0.092
0.092 * a^2 = 83^2 * a^2
0.092 = 83^2
a^2 = 0.092 / (83^2)
Taking the square root of both sides, we get:
a = sqrt(0.092 / (83^2))
Calculating this value, we find:
a ≈ 0.0177 km/h^2
So, the cheetah's average acceleration during the short sprint is approximately 0.0177 km/h^2.
(b) To find the displacement of the cheetah at t = 3.0 s, we'll use Equation 2 with the known values of initial velocity (u = 0 km/h), acceleration (a = 0.0177 km/h^2), and time (t = 3.0 s).
s = ut + (1/2)at^2
s = 0 * 3.0 + (1/2) * 0.0177 * (3.0^2)
Simplifying this expression, we find:
s ≈ 0.076 km
So, the displacement of the cheetah at t = 3.0 s is approximately 0.076 km.
(a) To determine the cheetah's average acceleration during the short sprint, we can use the following formula:
Acceleration (a) = (Final Speed - Initial Speed) / Time
In this case, the initial speed is 0 km/h (since the cheetah starts from rest), the final speed is 83 km/h, and the time is the time it takes for the cheetah to run the 46 meters.
First, let's convert the final speed from km/h to m/s:
Final Speed = 83 km/h = (83 km/h) * (1000 m/1 km) * (1 h/3600 s) = 23.05 m/s (rounded to two decimal places)
Now we can calculate the average acceleration:
Acceleration = (Final Speed - Initial Speed) / Time
Acceleration = (23.05 m/s - 0 m/s) / (46 m / 23.05 m/s)
Acceleration = 23.05 m/s / 2 s
Acceleration = 11.525 m/s^2
Therefore, the cheetah's average acceleration during the short sprint is approximately 11.525 m/s^2.
(b) To find the cheetah's displacement at t = 3.0 s, we can use the following equation of motion:
Displacement (d) = Initial Velocity * Time + (1/2) * Acceleration * Time^2
In this case, the initial velocity is 0 m/s, and the acceleration is the same as calculated in part (a).
First, let's calculate the displacement up to t = 3.0 s:
d = 0 * 3 + (1/2) * 11.525 * (3)^2
d = 0 + (1/2) * 11.525 * 9
d = 0 + 51.9125
d ≈ 51.91 m (rounded to two decimal places)
Therefore, the cheetah's displacement at t = 3.0 s is approximately 51.91 meters.