Question

The results for one patient show that the blood in the aorta begins at a speed of 0.10 m/s and undergoes constant acceleration for 38 ms, reaching a peak speed of 1.29 m/s. (a) What is the acceleration reflected in these data? (b) How far does the blood travel during this period?

Answer 1

0.002 m

Step-by-step explanation:

We can use the equations of motion to solve this problem.

(a) The initial velocity of the blood is u = 0.10 m/s, the final velocity is v = 1.29 m/s, the time taken is t = 38 ms = 0.038 s, and the acceleration is a (which is what we want to find). The equation that relates these quantities is:

v = u + at

Rearranging this equation, we get:

a = (v - u) / t = (1.29 - 0.10) / 0.038 = 33.68 m/s^2

Therefore, the acceleration of the blood is 33.68 m/s^2.

(b) To find the distance traveled by the blood during this period, we can use another equation of motion:

s = ut + (1/2)at^2

where s is the distance traveled. Substituting the values we have:

s = (0.10)(0.038) + (1/2)(33.68)(0.038)^2 = 0.002 m

Therefore, the blood travels a distance of 0.002 m during this period.

Answer 2

s ≈ 0.0603 meters

Explanation:

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

v = u + at

Where:

v = final velocity = 1.29 m/s

u = initial velocity = 0.10 m/s

a = acceleration (unknown)

t = time = 38 ms = 0.038 s

Rearranging the equation, we have:

a = (v - u) / t

Plugging in the values, we get:

a = (1.29 - 0.10) / 0.038

a = 1.19 / 0.038

a ≈ 31.32 m/s²

Therefore, the acceleration reflected in the data is approximately 31.32 m/s².

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

s = ut + (1/2)at²

Where:

s = distance traveled (unknown)

u = initial velocity = 0.10 m/s

t = time = 38 ms = 0.038 s

a = acceleration = 31.32 m/s²

Plugging in the values, we get:

s = (0.10 × 0.038) + (0.5 × 31.32 × 0.038²)

s ≈ 0.0038 + 0.0565

s ≈ 0.0603 meters

Therefore, the blood travels approximately 0.0603 meters during this period.