Answers:
1. Segment A (Position-Time on Ramp): The graph is a curve. On any graph that curves, the slope or steepness of the graph changes from one point on the graph to another. Since the slope is constantly changing, the velocity is non-constant. Also, as the graph lies above the x-axis and its slope is increasing, the velocity of the object is also increasing (speeding up) in a positive left direction.
Segment B (Position-Time on Flat Surface): The graph is a straight line, which represents constant velocity. The slope is negative, which indicates that the object is moving at a constant velocity left.
Including the part in which the car slows down to a stop, segment B would cross the x-axis with a curved line. Since it is a curved line, …show more content…
Acceleration on ramp: -0.70m/s^2 right
Acceleration on flat surface: 0m/s^2 right
No they are not similar. As the car rolls down the ramp, it accelerates and when the car is on a flat surface, it continues at a constant velocity (no acceleration). Therefore, it is entirely different as one has acceleration and the other does not.
However, if the segment continued until the ball came to a stop, the velocity would slowly decrease and then the acceleration on a flat surface in the right direction. Then, the difference between the accelerations would be that one is negative and the other is positive, meaning one is slowing down and one is increasing.
3. At one second, the instantaneous velocity of the car is, negative 0.869800824m/s right while the data from Logger Pro calculated the velocity of the car at one second to be negative 0.759364334m/s right. Therefore, the slope of the instantaneous velocity is steeper and therefore states the car accelerates more quickly than what the Logger Pro calculations believe. This deviation may have been caused since drawing a tangent is a very inaccurate as there are many other possible tangents that could be …show more content…
The hypothesis was that the type of motion an object has correlates to the slope of the path it travels on. In the graph produced in correspondence with the data, there were two portions, one to represent what happens on the ramp and one to represent what happens on a flat surface. On the flat surface, from 1.8 seconds to 2.67s, the slope using the line of best fit was zero in velocity-time. The acceleration also turned out to be zero metres per seconds squared, fitting perfectly in the hypothesis, the line had been horizontal (symmetrical to the x-axis) which indicates no change in slope or constant velocity. In the second portion, from 0.633s to 1.80s, the slope using the line of best fit was negative seven-tenths. This also directly corresponds to the acceleration in this portion, which is negative seven-tenths metre squared. Standing true to the hypothesis, since the line is diagonally facing downwards, the type of motion directly changes to uniform acceleration. Therefore, as the car is moving down the ramp, it is uniformly accelerating and when on a flat surface, it has constant
1. Segment A (Position-Time on Ramp): The graph is a curve. On any graph that curves, the slope or steepness of the graph changes from one point on the graph to another. Since the slope is constantly changing, the velocity is non-constant. Also, as the graph lies above the x-axis and its slope is increasing, the velocity of the object is also increasing (speeding up) in a positive left direction.
Segment B (Position-Time on Flat Surface): The graph is a straight line, which represents constant velocity. The slope is negative, which indicates that the object is moving at a constant velocity left.
Including the part in which the car slows down to a stop, segment B would cross the x-axis with a curved line. Since it is a curved line, …show more content…
Acceleration on ramp: -0.70m/s^2 right
Acceleration on flat surface: 0m/s^2 right
No they are not similar. As the car rolls down the ramp, it accelerates and when the car is on a flat surface, it continues at a constant velocity (no acceleration). Therefore, it is entirely different as one has acceleration and the other does not.
However, if the segment continued until the ball came to a stop, the velocity would slowly decrease and then the acceleration on a flat surface in the right direction. Then, the difference between the accelerations would be that one is negative and the other is positive, meaning one is slowing down and one is increasing.
3. At one second, the instantaneous velocity of the car is, negative 0.869800824m/s right while the data from Logger Pro calculated the velocity of the car at one second to be negative 0.759364334m/s right. Therefore, the slope of the instantaneous velocity is steeper and therefore states the car accelerates more quickly than what the Logger Pro calculations believe. This deviation may have been caused since drawing a tangent is a very inaccurate as there are many other possible tangents that could be …show more content…
The hypothesis was that the type of motion an object has correlates to the slope of the path it travels on. In the graph produced in correspondence with the data, there were two portions, one to represent what happens on the ramp and one to represent what happens on a flat surface. On the flat surface, from 1.8 seconds to 2.67s, the slope using the line of best fit was zero in velocity-time. The acceleration also turned out to be zero metres per seconds squared, fitting perfectly in the hypothesis, the line had been horizontal (symmetrical to the x-axis) which indicates no change in slope or constant velocity. In the second portion, from 0.633s to 1.80s, the slope using the line of best fit was negative seven-tenths. This also directly corresponds to the acceleration in this portion, which is negative seven-tenths metre squared. Standing true to the hypothesis, since the line is diagonally facing downwards, the type of motion directly changes to uniform acceleration. Therefore, as the car is moving down the ramp, it is uniformly accelerating and when on a flat surface, it has constant