The displacement of the Ariel Atom is 150 meters because it increases in velocity by 10 meters/sec/sec. You add up the 10 meters for every second and you get 150, 10+20+30+40+50. You use the graph to tell by adding 10 meters each second and it will create a parabola.
The displacement of the Arial atom is 150 meters. you can tell this because the every second the velocity increases by 10 m/s. It was moving for 5 seconds so the top speed was 50 m/s.Than, you just add the speeds together. 10+20+30+40+50=150.
The displacement of the Ariel Atom is 150 meters because it increases in velocity by 10 meters/sec/sec. Since it was 5 running for 5 seconds you would add up 10+20+30+40+50. You use the graph to plot the points and see how it changed from each second and it created a parabola.
The displacement is 127.5 meters when taken to the nearest tenth of a second. In one second the velocity will be 10 meters per second and the acceleration is constant at 10 meters per second per second. This means at the time of .1s the velocity will be 1 meter per second. Since the velocity is only 1 meter per second for about a tenth of a second the distance traveled is about a tenth of a meter. If you use this same concept for the velocity at .2s it will be 2 meters per second leaving the distance traveled as .2 meters. The distance traveled that you can calculate is only for that specific tenth of a second though since the velocity changes. I calculated the distance traveled for every tenth of a second up to the complete 5 seconds the Ariel Atom traveled and added them all up. A more specific answer could be arrived at since you could take the time to smaller parts thus making the velocity and distance traveled more specific but since numbers can be infinitely small there is an infinite amount of answers to this question depending on what you divide your time into.
The displacement of the Ariel Atom, when it travels for 5 seconds, is approximately 150 meters.
Note: If you assume that the car moved in a straight line on a flat surface, the distance traveled is equivalent to the displacement.
In order to come to this conclusion, you must understand that the car accelerates positively at 10 m/s/s. Since acceleration is equal to the change in Velocity over the change in Time (a=∆V/∆T), the car’s velocity increases by 10 m/s every second. For example, at P=0, before the car begins to move, its velocity is 0 m/s. At P= 1 second, the car’s velocity has increased 10 m/s. At P=2 seconds, the velocity has increased to 20 m/s. This happens consistently until the car reaches its maximum velocity or the car slows down.
Since the velocity changes, it is not possible to compute the distance traveled by multiplying the greatest velocity (50 m/s/s) by 5 seconds (the duration of the test), as the car was not moving at 50 m/s/s during the entire duration of the test. Instead, you must discern the distance traveled between each point and add them together.
At P=0 seconds, the car has not moved. The velocity is 0 m/s (P₀ =0 m). At P=1 second, the car has a velocity of 10 m/s. Therefore, from P=0 to P=1 it has moved 10 meters (P₀+P₁=10 m). At P=2 seconds, the car has a velocity of 20 m/s. From P=1 to P=2, it has moved 20 meters. From P=0 to P=2, it has moved 30 meters (P₀+P₁+P₂ =30 m). At P=3 seconds, the car has a velocity of 30 m/s. From P=2 to P=3, it has moved 30 meters. From P=0 to P=3, it has moved 60 meters (P₀+P₁+P₂+P₃ =60 m). At P=4 seconds, the car has a velocity of 40 m/s. From P=3 to P=4, it has moved 40 meters. From P=0 to P=4, it has moved 100 meters (P₀+P₁+P₂+P₃+P₄ =100 m). At P=5 seconds, the car has a velocity of 50 m/s. From P=1 to P=2, it has moved 50 meters. From P=0 to P=5, it has moved 150 meters (P₀+P₁+P₂+P₃+P₄+P₅ =150 m).
Therefore, since P₀+P₁+P₂+P₃+P₄+P₅ =150 m, the displacement is approximately 150 meters.
After doing further research, I have found that there are four kinematic equations that can be used to describe the motion of an object. There are two equations that are useful in this situation, but the most applicable one is d=vᵢ*t+½*a*t², where d is distance, vᵢ is initial velocity, t is time and a is acceleration.
In order to substitute our numbers into this equation we must discern what each of the variable represents in this case. Since the car isn't moving at P=0, the initial velocity is 0. The car is in motion for five seconds, so t=5. The acceleration is 10 m/s/s.
After substituting these values, the first equation becomes d= 0*5+½*10*5². d= 0*5+½*10*5² d= ½*10*25 d= 5*25 d=125
The displacement of the car is equal to 125 meters if you assume that the car is traveling on a flat surface in a straight line.
Even without using this formula, we know that the position must increase exponentially. The velocity is increasing at a constant rate (a=10 m/s/s), and the change in position over time is equivalent to the velocity. Therefore, the p-t graph must be parabolic. Without using the formula, we cannot discern the exact position at each interval, but you can still predict its relative location.
The displacment of the ariel atom is 150 meters, the velocity increases every second by 10 m/s. Because it was moving for 5 seconds, it would reach the top speed of 50 m/s. It wasn't 50m/s untill 5 seconds so you would have to add every other speed also. 10m/s:1s 20m/s:2s ect.. 10+20+30+40+50= 150
Sara VanDyke The position of the car starts at 0 meters and 0 seconds. The position is positive and increasing. The car accelerates in a positive and constant direction for about five seconds and reaches 100 mph. The velocity is constant and positive. At 0 seconds the velocity is 0m/hr. The car increases by 20 mph every second. 10m/s every second.
The displacement of the Arial atom is 150 meters. every second the velocity increases positively by 10 m/s. It was moving for 5 seconds so the top speed was 50 m/s.Then just add the speeds together. 10+20+30+40+50=150. AH
The displacement of the Ariel atom is 150 meters because if it's velocity was increasing by 10 m/s/s for 5 seconds then the question would look like so: 10+20+30+40+50. I'm posting this so late because I couldn't reach a computer at home so Im using my phone.
The displacement of the Ariel Atom is 150 meters because it increases in velocity by 10 meters/sec/sec. You add up the 10 meters for every second and you get 150, 10+20+30+40+50. You use the graph to tell by adding 10 meters each second and it will create a parabola.
ReplyDeleteThe displacement of the Arial atom is 150 meters. you can tell this because the every second the velocity increases by 10 m/s. It was moving for 5 seconds so the top speed was 50 m/s.Than, you just add the speeds together. 10+20+30+40+50=150.
ReplyDeleteThe displacement of the Ariel Atom is 150 meters because it increases in velocity by 10 meters/sec/sec. Since it was 5 running for 5 seconds you would add up 10+20+30+40+50. You use the graph to plot the points and see how it changed from each second and it created a parabola.
ReplyDeleteThe displacement is 127.5 meters when taken to the nearest tenth of a second. In one second the velocity will be 10 meters per second and the acceleration is constant at 10 meters per second per second. This means at the time of .1s the velocity will be 1 meter per second. Since the velocity is only 1 meter per second for about a tenth of a second the distance traveled is about a tenth of a meter. If you use this same concept for the velocity at .2s it will be 2 meters per second leaving the distance traveled as .2 meters. The distance traveled that you can calculate is only for that specific tenth of a second though since the velocity changes. I calculated the distance traveled for every tenth of a second up to the complete 5 seconds the Ariel Atom traveled and added them all up. A more specific answer could be arrived at since you could take the time to smaller parts thus making the velocity and distance traveled more specific but since numbers can be infinitely small there is an infinite amount of answers to this question depending on what you divide your time into.
ReplyDeleteThe displacement of the Ariel Atom, when it travels for 5 seconds, is approximately 150 meters.
ReplyDeleteNote: If you assume that the car moved in a straight line on a flat surface, the distance traveled is equivalent to the displacement.
In order to come to this conclusion, you must understand that the car accelerates positively at 10 m/s/s. Since acceleration is equal to the change in Velocity over the change in Time (a=∆V/∆T), the car’s velocity increases by 10 m/s every second.
For example, at P=0, before the car begins to move, its velocity is 0 m/s. At P= 1 second, the car’s velocity has increased 10 m/s. At P=2 seconds, the velocity has increased to 20 m/s. This happens consistently until the car reaches its maximum velocity or the car slows down.
Since the velocity changes, it is not possible to compute the distance traveled by multiplying the greatest velocity (50 m/s/s) by 5 seconds (the duration of the test), as the car was not moving at 50 m/s/s during the entire duration of the test. Instead, you must discern the distance traveled between each point and add them together.
At P=0 seconds, the car has not moved. The velocity is 0 m/s (P₀ =0 m).
At P=1 second, the car has a velocity of 10 m/s. Therefore, from P=0 to P=1 it has moved 10 meters (P₀+P₁=10 m).
At P=2 seconds, the car has a velocity of 20 m/s. From P=1 to P=2, it has moved 20 meters. From P=0 to P=2, it has moved 30 meters (P₀+P₁+P₂ =30 m).
At P=3 seconds, the car has a velocity of 30 m/s. From P=2 to P=3, it has moved 30 meters. From P=0 to P=3, it has moved 60 meters (P₀+P₁+P₂+P₃ =60 m).
At P=4 seconds, the car has a velocity of 40 m/s. From P=3 to P=4, it has moved 40 meters. From P=0 to P=4, it has moved 100 meters (P₀+P₁+P₂+P₃+P₄ =100 m).
At P=5 seconds, the car has a velocity of 50 m/s. From P=1 to P=2, it has moved 50 meters. From P=0 to P=5, it has moved 150 meters (P₀+P₁+P₂+P₃+P₄+P₅ =150 m).
Therefore, since P₀+P₁+P₂+P₃+P₄+P₅ =150 m, the displacement is approximately 150 meters.
TR
After doing further research, I have found that there are four kinematic equations that can be used to describe the motion of an object. There are two equations that are useful in this situation, but the most applicable one is d=vᵢ*t+½*a*t², where d is distance, vᵢ is initial velocity, t is time and a is acceleration.
ReplyDeleteIn order to substitute our numbers into this equation we must discern what each of the variable represents in this case. Since the car isn't moving at P=0, the initial velocity is 0. The car is in motion for five seconds, so t=5. The acceleration is 10 m/s/s.
After substituting these values, the first equation becomes d= 0*5+½*10*5².
d= 0*5+½*10*5²
d= ½*10*25
d= 5*25
d=125
The displacement of the car is equal to 125 meters if you assume that the car is traveling on a flat surface in a straight line.
Even without using this formula, we know that the position must increase exponentially. The velocity is increasing at a constant rate (a=10 m/s/s), and the change in position over time is equivalent to the velocity. Therefore, the p-t graph must be parabolic. Without using the formula, we cannot discern the exact position at each interval, but you can still predict its relative location.
TR
The displacment of the ariel atom is 150 meters, the velocity increases every second by 10 m/s. Because it was moving for 5 seconds, it would reach the top speed of 50 m/s. It wasn't 50m/s untill 5 seconds so you would have to add every other speed also. 10m/s:1s 20m/s:2s ect.. 10+20+30+40+50= 150
ReplyDeleteSara VanDyke
ReplyDeleteThe position of the car starts at 0 meters and 0 seconds. The position is positive and increasing. The car accelerates in a positive and constant direction for about five seconds and reaches 100 mph. The velocity is constant and positive. At 0 seconds the velocity is 0m/hr. The car increases by 20 mph every second. 10m/s every second.
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ReplyDeletelol day 69
ReplyDeleteThe displacement of the Arial atom is 150 meters. every second the velocity increases positively by 10 m/s. It was moving for 5 seconds so the top speed was 50 m/s.Then just add the speeds together. 10+20+30+40+50=150.
ReplyDeleteAH
The displacement of the Ariel atom is 150 meters because if it's velocity was increasing by 10 m/s/s for 5 seconds then the question would look like so: 10+20+30+40+50. I'm posting this so late because I couldn't reach a computer at home so Im using my phone.
ReplyDelete150 idk
ReplyDelete