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Determine the magnitude of the acceleration for the speeding up phase

November 14, 2021 by moyways Leave a Comment

a. Determine the magnitude of acceleration during the speeding up phase.

Include two significant figures and the appropriate units with your answer.

b. Calculate the magnitude of the acceleration during the slowing down phase.

Include the appropriate units and two significant figures in your response.

2 Answers

(Change in velocity) * (Change in time) = Acceleration

vf – v1 = change in velocity

Time change = tf – ti

The acceleration is equal to (vf – v1) * (tf – ti)

As a rule of thumb, you measure the horizontal distance between the y axis and the first vertical line as well as the vertical distance between the x axis and the first horizontal line in order to determine the values of velocity and time. 0.2 seconds is the horizontal distance. 0.5 m/s is the vertical distance.

The graph was copied and pasted into a Microsoft Word document. Using a ruler, I measured the horizontal distance from the y axis to the 1st vertical line and the vertical distance from the x axis to the 1st horizontal line.

The distance between the two points is approximately 3.4 cm. There are 3.4 cm wide and 3.4 cm tall squares between the two points.

Determine the magnitude of the acceleration for the speeding up phase.?
Determine the magnitude of the acceleration for the speeding up phase.?

Using a ruler, I measured the horizontal distance from the y axis to the highest position and the vertical distance from the x axis to the highest point.

This is the horizontal distance of 4.3 cm

Approximately 5.5 cm is the vertical distance

To determine the actual time and velocity at the highest point, set up a proportion.

A ratio of 3.4 to 4.3 is 0.2/t, and time is 0.253 seconds

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In the example given, 3.4 / 5.5 = 0.5 v, Velocity = 0.89 m/s

At 0.253 seconds, the velocity of the blood is 0.809 m/s

At 0.2 seconds and 0.4 seconds, the velocity of the blood is 0 m/s

For the speeding up phase, acceleration = (0.809 – 0) ÷ (0.253 – 0.2) ≈ 15.3 m/s^2

For the slowing down phase, acceleration = (0 – 0.809 ÷ (0.4 – 0.253) ≈ -5.5 m/s^2

Based on my measurements, these values have been calculated. Use the exact time and velocity from the highest point of the graph if you know them!

OR

You measure !

The answers are 11 m/s2

and 5.6 m/s2

i just got it on mastering physics

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