a cheetah can accelerate from rest to 24 m/s in 2.0 s . Assuming the acceleration is constant over the time interval, a)what is the magnitude of the acceleration of the cheetah? b) what is the distance traveled by the cheetah in these 2.0 s.? c) a runner can accelerate from rest to 6.0 m/s in the same time , 2.0 s What id the magnitude of the acceleration of the runner? By what factor is the cheetah's average acceleration magnitude greater than that of the runner?

a. a = (Vt - Vo) / t = (24 m/s - 0 m/s) / 2 s = 24 / 2 = 12 m/s^2.

b. d = 0.5at^2 = 0.5 * 12 m/s^2 * (2s)^2 = 0.5 * 12 * 4 = 24 m.

c. a = (6 m/s - 0 m/s) / 2 s 3 m/s^2

12 m/s^2 * ( 1 / 3 )s^2/m = 4.

thank you so much

a) To find the magnitude of acceleration, we can use the equation:

acceleration = (final velocity - initial velocity) / time

Given that the initial velocity (u) is 0 m/s, the final velocity (v) is 24 m/s, and the time (t) is 2.0 s, we can substitute these values into the formula:

acceleration = (24 m/s - 0 m/s) / 2.0 s
acceleration = 24 m/s / 2.0 s
acceleration = 12 m/s²

Therefore, the magnitude of the acceleration of the cheetah is 12 m/s².

b) To find the distance traveled by the cheetah, we can use the equation of motion:

distance = initial velocity × time + (1/2) × acceleration × time²

Again, we know that the initial velocity (u) is 0 m/s, the time (t) is 2.0 s, and the acceleration is 12 m/s². Substituting these values into the equation:

distance = 0 m/s × 2.0 s + (1/2) × 12 m/s² × (2.0 s)²
distance = 0 m + 6 m/s² × 4 s²
distance = 0 m + 24 m
distance = 24 m

Therefore, the distance traveled by the cheetah in these 2.0 s is 24 meters.

c) Using the same formula as in part a, we can find the magnitude of the runner's acceleration:

acceleration = (final velocity - initial velocity) / time
acceleration = (6 m/s - 0 m/s) / 2.0 s
acceleration = 6 m/s / 2.0 s
acceleration = 3 m/s²

Therefore, the magnitude of the acceleration of the runner is 3 m/s².

To find the factor by which the cheetah's average acceleration magnitude is greater than that of the runner, we divide the cheetah's acceleration magnitude by the runner's acceleration magnitude:

factor = acceleration of cheetah / acceleration of runner
factor = 12 m/s² / 3 m/s²
factor = 4

Therefore, the cheetah's average acceleration magnitude is 4 times greater than that of the runner.

To answer these questions, we can use the equations of motion.

a) The magnitude of acceleration (a) can be calculated using the equation:

a = (v - u) / t

Where:
v = final velocity = 24 m/s
u = initial velocity = 0 m/s (since the cheetah starts from rest)
t = time = 2.0 s

Substituting the given values:

a = (24 m/s - 0 m/s) / 2.0 s
a = 12 m/s²

Therefore, the magnitude of the acceleration of the cheetah is 12 m/s².

b) The distance traveled by the cheetah can be calculated using the equation:

s = ut + (1/2)at²

Where:
s = distance traveled
u = initial velocity = 0 m/s
t = time = 2.0 s
a = acceleration = 12 m/s²

Substituting the given values:

s = (0 m/s)(2.0 s) + (1/2)(12 m/s²)(2.0 s)²
s = 0 m + 12 m/s² * 2.0 s²
s = 12 m/s² * 4.0 s²
s = 48 m

Therefore, the cheetah would travel a distance of 48 meters in these 2.0 seconds.

c) The magnitude of acceleration for the runner can be calculated in the same way as the cheetah:

a = (v - u) / t

Where:
v = final velocity = 6.0 m/s
u = initial velocity = 0 m/s (since the runner starts from rest)
t = time = 2.0 s

Substituting the given values:

a = (6.0 m/s - 0 m/s) / 2.0 s
a = 3.0 m/s²

Therefore, the magnitude of the acceleration of the runner is 3.0 m/s².

To find the factor by which the cheetah's average acceleration magnitude is greater than that of the runner, we can calculate the ratio:

Factor = cheetah's acceleration / runner's acceleration
Factor = 12 m/s² / 3.0 m/s²
Factor = 4

Therefore, the cheetah's average acceleration magnitude is four times greater than that of the runner.