Type your question here 3. Complete speedometer positions and odometer readings
ID: 2232704 • Letter: T
Question
Type your question here
3. Complete speedometer positions and odometer readings in Figure 3.9 Then complete the chart below.
time (s)
speed (m/s)
distance (m)
0
0
0
1
10
5
2
20
20
3
4
5
6
4. Imagine that an astronaut on a distant planet drops a rock from a 80 m tall cliff. The rock strikes th ground 2 seconds later. What is the value of gravitational acceleration there?
5. Analyze Figure 3.8. Now imagine that you find yourself on a planet X. You toss up a rock with initial speed of 20 m/s. The table below shows few velocities and positions of the rock. Complete all entries for velocites and positions. Pay attention to signs. What is the value of gravitational acceleration on the planet X?
time (s)
velocity (m/s)
distance (m)
0
20
0
1
15
17.5
2
10
3
4
5
6
7
8
time (s)
speed (m/s)
distance (m)
0
0
0
1
10
5
2
20
20
3
4
5
6
Explanation / Answer
3. You can see that the speed is increasing at a constant rate. First it's zero, then it gains 10 m/s in 1s, then it gains another 10 m/s in another second, etc. This means that the object is gaining speed at a rate of 10 m/s per second, ie it has an acceleration of 10 m/s^2. So, with each consecutive second, it will gain another 10 m/s. This means that at 3s, it will be moving at 30 m/s, at 4, 40, at 5, 50, at 6, 60. Now, the distance is another story. The equation for distance assuming that you start from rest is
(d = (1/2)at^{2})
So, at 3 seconds, it's
(d = (1/2)(10 m/s^{2})(3s)^2 = 45m)
(d = (1/2)(10 m/s^{2})(4s)^{2} = 80m)
At 5s
(d = (1/2)(10 m/s^{2})(5s)^{2} = 125m)
At 6s
(d = (1/2)(10 m/s^{2})(6s)^{2} = 180m)
4. We will use the same equation for as before
(d = (1/2)at^{2})
where the distance travelled is 80m, the time is 2 seconds, and the acceleration is the unknown gravitational acceleration:
(a = 2d/t^{2} = 2(80m)/(2s)^{2} = 40 m/s^{2})
5. In this scenario, the velocity is decreasing by 5 m/s per second, so the acceleration is -5m/s^2, and the velocity at 3s will be 5 m/s, at 4, 0 m/s, at 5, -5, at 6, -10, at 7, -15, and at 8s, -20 m/s. Now, to find the distances. The equation is basically the same, but there is an initial velocity, so it has an extra term in it, v_0t:
At 2s,
(d = v_{0}t + (1/2)at^{2} = (20 m/s)(2s) + (1/2)(- 5m/s^{2})(2s)^{2} = 30m)
At 3s,
(d = v_{0}t + (1/2)at^{2} = (20 m/s)(3s) + (1/2)(- 5m/s^{2})(3s)^{2} = 37.5m)
At 4s,
(d = (20 m/s)(4s) + (1/2)(- 5m/s^{2})(4s)^{2} = 40m)
At 5s,
(d = (20 m/s)(5s) + (1/2)(- 5m/s^{2})(5s)^{2} = 37.5 m)
At 6s,
(d = (20 m/s)(6s) + (1/2)(- 5m/s^{2})(6s)^{2} = 30m)
At 7s,
(d = (20 m/s)(7s) + (1/2)(- 5m/s^{2})(7s)^{2} = 17.5m)
And at 8s,
(d = (20 m/s)(8s) + (1/2)(- 5m/s^{2})(8s)^{2} = 0m)
Hope this helps
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