Kaplunovsky - hw 02 - homework 02 Due: Jan 31 2007, 4:00 pm...

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Unformatted text preview: homework 02 Due: Jan 31 2007, 4:00 pm 1 Question 1, chap 2, sect 3. part 1 of 4 5 points The position versus time for a certain ob- ject moving along the x-axis is shown. The objects initial position is 3 m. 4 2 2 4 6 1 2 3 4 5 6 7 8 9 b b b b b b time (s) position(m) Find the instantaneous velocity at 0 . 5 s. Correct answer: 7 m / s (tolerance 1 %). Explanation: The instantaneous velocity is the slope of the tangent line at that point. v = (4 m) ( 3 m) (1 s) (0 s) = 7 m / s . Question 2, chap 2, sect 3. part 2 of 4 5 points Find the instantaneous velocity at 2 . 5 s. Correct answer: . 666667 m / s (tolerance 1 %). Explanation: v = (2 m) (4 m) (4 s) (1 s) = . 666667 m / s . Question 3, chap 2, sect 3. part 3 of 4 5 points Find the instantaneous velocity at 4 . 5 s. Correct answer: 0 m / s (tolerance 1 %). Explanation: v = (2 m) (2 m) (5 s) (4 s) = 0 m / s . Question 4, chap 2, sect 3. part 4 of 4 5 points Find the instantaneous velocity at 8 s. Correct answer: 1 m / s (tolerance 1 %). Explanation: v = (0 m) ( 2 m) (9 s) (7 s) = 1 m / s . Question 5, chap 2, sect 2. part 1 of 3 10 points Consider the plot below describing motion along a straight line with an initial position of x = 10 m. 2 1 1 2 3 4 5 1 2 3 4 5 6 7 8 9 b b b b b time (s) velocity(m/s) What is the position at 2 seconds? Correct answer: 15 m (tolerance 1 %). Explanation: The initial position given in the problem is 10 m. b b b b b b b b 1 2 3 4 5 6 7 8 9 1 2 3 4 5 1 2 time (s) velocity(m/s) homework 02 Due: Jan 31 2007, 4:00 pm 2 The position at 2 seconds is 10 meters plus the area of the triangle (shown in gray in the above plot) x = (10 m) + 1 2 [(2 s) (0 s)] [(5 m / s) (0 m / s)] = 15 m ; however, it can also be calculated: x = x i + v i ( t f t i ) + 1 2 ( t f t i ) 2 = (10 m) + (0 m / s) [(2 s) (0 s)] + 1 2 (2 . 5 m / s 2 ) [(2 s) (0 s)] 2 = 15 m . Question 6, chap 2, sect 2. part 2 of 3 10 points What is the position at 6 seconds? Correct answer: 29 m (tolerance 1 %). Explanation: The position is 15 m plus the area of the trapezoid x = (15 m) + 1 2 [(6 s) (2 s)] [(2 m / s) + (5 m / s)] = 29 m ; however, it can also be calculated: x = x i + v i ( t f t i ) + 1 2 ( t f t i ) 2 = (15 m) + (5 m / s) [(6 s) (2 s)] + 1 2 ( . 75 m / s 2 ) [(6 s) (2 s)] 2 = 29 m , where the position is 29 m plus the area of the trapezoid: one-half the base [(6 s) (2 s)] the sum of the heights [(2 m / s) + (5 m / s)]. Question 7, chap 2, sect 2. part 3 of 3 10 points What is the position at 8 seconds? Correct answer: 27 . 6667 m (tolerance 1 %)....
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Kaplunovsky - hw 02 - homework 02 Due: Jan 31 2007, 4:00 pm...

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