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08/26/09 - Chapter 2 - Describing Motion (Part II) |
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;){kind=small} | Constant velocity: |
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;){kind=small} | Positive velocity: |
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;){kind=small} | Velocity-time graph |
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;){kind=small} | Position-time graph with initial position = 0 |
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;){kind=small} | Position-time graph with positive initial position |
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;){kind=small} | Position-time graph with negative initial position |
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;){kind=small} | Negative velocity: |
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;){kind=small} | Velocity-time graph |
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;){kind=small} | Position-time graph with initial position = 0 |
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;){kind=small} | Position-time graph with positive initial position |
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;){kind=small} | Position-time graph with negative initial position |
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;){kind=small} | General formula for position as a function of time with constant velocity v: |
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;){kind=small} | Problems involving constant velocity: |
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| A car starts 10 miles south of a city and drives north at 45 miles/hour for 20 minutes and then stops. Where is the car when it stops? Sketch graphs of the position of the car versus time and the velocity of the car versus time. |
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| A red car starts 15 miles west of town and drives east at a speed of 30 miles/hour. A blue car starts 25 miles west of town and drives east at 40 miles/hour. At what time will the blue car catch up with the red car? Where are they when the blue car catches up with the red car? |
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;){kind=small} | Constant acceleration: |
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;){kind=small} | Example 1: |
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;){kind=small} | For an object changing its velocity at a constant rate of 2 m/s every second (constant acceleration of 2 m/s2), make a table showing time, acceleration, instantaneous velocity, average velocity and displacement. Draw the a-t, v-t and d-t graphs. |
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;){kind=small} | Table |
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;){kind=small} | a-t graph |
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;){kind=small} | v-t graph |
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;){kind=small} | d-t graph |
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;){kind=small} | Positive acceleration |
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;){kind=small} | Acceleration-time graph |
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;){kind=small} | Velocity-time graph with initial velocity = 0 |
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;){kind=small} | Velocity-time graph with positive initial velocity |
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;){kind=small} | Velocity-time graph with negative initial velocity |
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;){kind=small} | Position-time graph with initial velocity = 0 and initial position = 0 |
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;){kind=small} | Position-time graph with positive initial velocity and initial position = 0 |
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;){kind=small} | Position-time graph with negative initial velocity and initial position = 0 |
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;){kind=small} | Position-time graph with initial velocity = 0 and positive initial position |
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;){kind=small} | Position-time graph with positive initial velocity and positive initial position |
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;){kind=small} | Position-time graph with negative initial velocity and positive initial position |
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;){kind=small} | Position-time graph with initial velocity = 0 and negative initial position |
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;){kind=small} | Position-time graph with positive initial velocity and negative initial position |
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;){kind=small} | Position-time graph with negative initial velocity and negative initial position |
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;){kind=small} | Negative acceleration |
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;){kind=small} | Acceleration-time graph |
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;){kind=small} | Velocity-time graph with initial velocity = 0 |
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;){kind=small} | Velocity-time graph with positive initial velocity |
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;){kind=small} | Velocity-time graph with negative initial velocity |
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;){kind=small} | Position-time graph with initial velocity = 0 and initial position = 0 |
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;){kind=small} | Position-time graph with positive initial velocity and initial position = 0 |
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;){kind=small} | Position-time graph with negative initial velocity and initial position = 0 |
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;){kind=small} | Position-time graph with initial velocity = 0 and positive initial position |
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;){kind=small} | Position-time graph with positive initial velocity and positive initial position |
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;){kind=small} | Position-time graph with negative initial velocity and positive initial position |
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;){kind=small} | Position-time graph with initial velocity = 0 and negative initial position |
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;){kind=small} | Position-time graph with positive initial velocity and negative initial position |
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;){kind=small} | Position-time graph with negative initial velocity and negative initial position |
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;){kind=small} | General formula for velocity as a function of time with constant acceleration |
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;){kind=small} | General formula for position as a function of time with constant acceleration |
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;){kind=small} | Problems involving constant acceleration: |
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;){kind=small} | Example 2: |
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| A car accelerates at a constant rate from 10 m/s to 30 m/s in 5 seconds. Sketch the velocity-time graph, the acceleration-time graph and the displacement-time graph. |
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;){kind=small} | Table |
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;){kind=small} | v-t graph |
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;){kind=small} | a-t graph |
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;){kind=small} | d-t graph |
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;){kind=small} | Example 3: |
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;){kind=small} | In coming to a stop, a car leaves skid marks 92 m long on the highway. Assuming a deceleration of 7 m/s2 (negative acceleration of -7 (m/s)/s), sketch the velocity-time, acceleration-time and displacement-time graphs and estimate the speed of the car just before braking. |
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;){kind=small} | v-t graph |
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;){kind=small} | a-t graph |
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;){kind=small} | d-t graph |
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| From the sketch of the v-t graph, we can get the equation of the velocity as a function of time: v = -7t + v0. This tells us the car will stop when v = 0 = -7t + v0, so the time it takes to stop is t = v0/7. We can also see from the sketch that the area under the v-t graph is the distance traveled: 92m. This tells us that (1/2)v0t = 92. Substituting t = v0/7, tells us (1/2)v0(v0/7) = 92, or v02 = 1288, so v0 = 36 m/s. |
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;){kind=small} | Example 4: |
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| Sketch v-t, a-t and d-t graphs to determine the stopping distances for a car with an initial speed of 80 km/h and human reaction time of 1.0 s, for an acceleration of (a) -4 m/s2; (b) -8 m/s2. |
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;){kind=small} | (a) acceleration = -4 m/s2: |
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;){kind=small} | v-t graph |
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| From the v-t graph we can see that the car travels 22.2 m/s for the first second, and then starts to slow down at the rate of 4 m/s every second. During the first second it travels the distance d1 = 22.2 m/s * 1 s = 22.2 m. It then loses velocity at the rate of 4 m/s every second, so ∆v = -22.2 m/s = -4 m/s/s * ∆t. This tells us ∆t = 22.2/4 = 5.55 s. During that time the car travels the distance d2 = (1/2)(22.2 m/s)(5.55 s) = 61.6 m. So, the total stopping distance is d1 + d2 = 22.2 m + 61.6 m = 83.8 m = 84 m. |
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;){kind=small} | a-t graph |
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;){kind=small} | d-t graph |
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;){kind=small} | (b) acceleration = -8 m/s2: |
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;){kind=small} | v-t graph |
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| From the v-t graph we can see that the car travels 22.2 m/s for the first second, and then starts to slow down at the rate of 4 m/s every second. During the first second it travels the distance d1 = 22.2 m/s * 1 s = 22.2 m. It then loses velocity at the rate of 8 m/s every second, so ∆v = -22.2 m/s = -8 m/s/s * ∆t. This tells us ∆t = 22.2/8 = 2.78 s. During that time the car travels the distance d2 = (1/2)(22.2 m/s)(2.78 s) = 30.9 m. So, the total stopping distance is d1 + d2 = 22.2 m + 30.9 m = 53.1 m = 53 m. |
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;){kind=small} | a-t graph |
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;){kind=small} | d-t graph |
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;){kind=small} | Example 5: |
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| A stone is dropped from the top of a cliff. It hits the ground below after 5.5 s. Sketch the v-t, a-t and d-t graphs assuming the acceleration of gravity is 9.8 m/s/s and air-resistance is negligible. How high is the cliff? |
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;){kind=small} | v-t graph |
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| From the v-t graph we can see that the equation for the velocity curve is v = -9.8t. The area between the velocity curve and the t-axis is the distance fallen, which is the height of the cliff. The velocity when the rock hits the ground is v = -9.8 m/s/s (5.5 s) = 53.9 m/s. Then the area between the velocity curve and the t-axis is h = (1/2)(53.9 m/s)(5.5 s) = 148 m. So the height of the cliff is about 150 m. |
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;){kind=small} | a-t graph |
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;){kind=small} | d-t graph |
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;){kind=small} | Example 6: |
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| A stone is thrown vertically upward with a speed of 28 m/s. Sketch the v-t, a-t and d-t graphs to determine (a) how fast it is moving when it reaches a height of 15 m; (b) how much time is required to reach this height; (c) why there are two answers to part (b). |
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;){kind=small} | v-t graph |
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| From the v-t graph we can see that the equation for the velocity is v = -9.8t + 28. The stone stops when the curve crosses the t-axis v = 0 = -9.8t + 28. This corresponds to the peak height of the stone. So, the stone reaches its peak height at t = (28 m/s)/(9.8 m/s/s) = 2.86 s. Therefore the peak height is (1/2)(28 m/s)(2.86 s) = 40.0 m. So the stone will reach a height of 15 m once on the way up and again on the way down. To find the time when it reaches a height of 15 m, we look at the v-t graph below: |
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| We want to find the time t1 when the area under the v-t graph is 15 m. The area of the trapezoid is 15 = ((28 + (-9.8t1 + 28))/2)t1 = -4.9t12 + 28t1. This is a quadratic equation: -4.9t12 + 28t1 - 15 = 0. The solution is: t1 = 0.60 s and t1 = 5.1 s. The stone is 15 m above the ground twice: once at 0.60 s on the way up and again at 5.1 s on the way down. |
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;){kind=small} | Homework |
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| Read the first two sections of Chapter 2, pp. 26-38. |
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| Answer the questions in HW2 on WebAssign: http://webassign.net/login.html |
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