Suggested Problems Solutions

# Suggested Problems Solutions - SUGGESTED PROBLEMS P2.6...

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SUGGESTED PROBLEMS P2.6. Prepare: The slope of the position graph is the velocity graph. Solve: (a) Looking at the position graph we see the ball start at 6m and go up to 6m so that must be where the children are. (b) Assess: We can see that the first child rolled the ball faster to the second child than the second child did back to the first. P2.7. Prepare: To get a position from a velocity graph we count the area under the curve. Solve: (a) (b) We need to count the area under the velocity graph (area below the x -axis is subtracted). There are 18 m of area above the axis and 4 m of area below. 18m 4m =14m. Assess: These numbers seem reasonable; a mail carrier could back up 4 m. It is also important that the problem state what the position is at t = 0, or we wouldn’t know how high to draw the position graph. P2.8. Prepare: To get a position from a velocity graph we count the area under the curve. Solve: (a)

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(b) We need to count the area under the velocity graph (area below the x -axis is subtracted). There are 12 m of area below the axis and 12 m of area above. 12m 12m = 0m. (c) A football player runs left at 3 m/s for 4 s, then cuts back to the right at 3 m/s for 2 s, then walks (continuing to the right) back to the starting position. Assess: We note an abrupt change of velocity from 3 m/s left to 3 m/s right at 4 s. It is also important that the problem state what the position is at t =0, or we wouldn’t know how high to draw the position graph. P2.17. Prepare: The graph in Figure P2.17 shows distinct slopes in the time intervals: 0 – 1s, 1 s – 2 s, and 2 s – 4 s. We can thus obtain the velocity values from this graph using v = x / t . Solve: (a) (b) There is only one turning point. At t = 2 s the velocity changes from +20 m/s to 10 m/s, thus reversing the direction of motion. At t = 1 s, there is an abrupt change in motion from rest to +20 m/s, but there is no reversal in motion. Assess: As shown above in (a) , a positive slope must give a positive velocity and a negative slope must yield a negative velocity. P2.18. Prepare: We need to estimate the distance from the heart to the brain for the second part. 30 cm is a good estimate. Solve: (a) The distance traveled is the area under the v y graph. The area of a triangle is 1 2 BH . y =area= 1 2 BH = 1 2 (0.20 s)(0.75 m/s) = 7.5 cm (b) t = y v y = 30 cm 7.5 cm/beat =4.0 beats Assess: Four beats seems reasonable. P2.19. Prepare: Please refer to Figure P2.19. Since displacement is equal to the area under the velocity graph between t i and t f , we can find the car’s final position from its initial position and the area.
Solve: (a) Using the equation x f = x i + area of the velocity graph between t i and t f , x 2 s 10 m area of trapezoid between 0 s and 2 s 1 2 (12 m/s 4 m/s)(2 s) 26 m x 3 s area of triangle between 0 s and 3 s 1 2 (12 m/s)(3 s) 28 m x 4 s x 3 s area between 3 s and 4 s 1 2 ( 4 m/s)(1 s) (b)

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Suggested Problems Solutions - SUGGESTED PROBLEMS P2.6...

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