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Discussions:

1. Describe the position versus time plot of the Graph screen. Why does the distancebegin at a maximum and decrease as the cart moves up the inclined plane?

The position versus time graph is plotted on as a positive parabola; it starts at

the maximum and reaches the minimum and goes back to maximum. The position timegraph shoes the distance away from the reference point. In this case, the reference point is

the sensor. The distance at the starting point, is the furthest distance away from the

sensor, therefore making the number the largest (or the maximum). As the cart goes upthe incline plane, the distance is shorter, eventually reaching the minimum. This occurs

because the cart will reach its potential distance before rolling back down the incline

plane due to acceleration. As it rolls down, it will meet the same distance as when the cartrolled up.

2. Choose three spots from your sketch of displacement time graph, half way up,middle, and half way down. Draw the tangent lines and calculate their slopes. Please

show your calculation. What do these slopes represent?Choose [down] as +

Half way up: (0.9,0.2); (0.3,0.45)

Vinst= slope=

=-0.417m/s

= 0.417 m/s [up]

Middle: (0.5,0.4); (1.2,0.4)

Vinst= slope=

=0

Half way down: (1.0,0.2); (1.5,0.375)

Vinst= slope=

=0.35 m/s [down]

The slopes represent the instantaneous velocity

at the given points by creating a tangent line,

and calculating the slopes. However, eachinstantaneous velocity tells something different.

The instantaneous velocity for half way up shows

the velocity going up, which is actually a negative

slope. It is a negative slope because it is starting below the reference point, but as it goesup to the reference point, it is getting closer. Therefore, the position versus time graph

will start at a maximum furthest away from the reference point, but at it moves up the

incline plane, it gets closer the reference point (sensor) therefore being at a closerdistance. This is the reason why the velocity on the way up would be negative. During the

instantaneous velocity at the middle, it would result to zero. This would occur because

during the middle, the cart would reach its potential, making the velocity to a complete

stop at a very split second. The tangent line created was a straight horizontal line, parallel

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to the x-axis therefore making the slope zero. The instantaneous velocity for half way

down shows the velocity of the cart going back down the incline plane. This is a positive

slope because it is starting at the close to the reference point, but then going away. As it

goes down, it is further and further away from the reference point; eventually making itback to the maximum; therefore making the slope on the way down, positive.

3. Describe the velocity versus time plot. Should this graph be a straight line? Why orwhy not?The velocity versus time graph is plotted as a positive slope starting from the 4

th

quadrant. The positive slope is not 100% linear due to the imperfections of the human

force. However, the slope is roughly a linear line. This makes the y-axis value (velocity)

start as negative. The motion towards the sensor as negative because it is ratio betweenthe position and time is a negative slope therefore making the velocity negative for a

given period of time. However, the negative velocity does not actually represent anegative speed; it represents the cart going up the incline plane. In this case the positive

slope is going down the incline plane. At one point the velocity reaches 0, this occursbecause when the cart reaches its potential distance, the cart stops for a split second. This

graph should be a straight line because the velocity is continuously increasing even

though it is going in different direction. The velocity is increasing towards one direction,therefore creating a positive slope (positive acceleration). This positive acceleration

means that the acceleration is accelerating downwards.

4. Compare the calculations in discussion #2 to the data in velocity time graph, do theyagree?

The calculations in #2 do agree with the graph. It agrees with the graph because

while going up the incline plane, the velocity is negative, and during the given time thatits going up, the velocity during the same time is negative. The calculation of theinstantaneous velocity also matches the velocity on the graph for the given time. At the

middle point, the velocity does reach 0. After the velocity of 0, the velocity becomes

positive (which is when the cart is rolling back down the plane). The velocity going downis positive, and the instantaneous velocity during half of going down also matches the

velocity for the given time.

5. Describe the acceleration versus time plot of the Graph display. Should this graphbe a horizontal line? Why or why not?The acceleration versus time graph is roughly a horizontal line with a slope of zero. It

ranges from 0.8 to 1.4. Even though it should be a constant acceleration, due to human

error, and the sources of error, the numbers vary by a little. The reason that it should be ahorizontal line is because the velocity increases, but the change in velocity over the timeperiod is the acceleration. Therefore making the ratio between the velocity and time the

same at each time therefore making the acceleration the same number.

6. How does the acceleration determined in the plot of velocity compare to the averagevalue of acceleration from the plot of acceleration?The acceleration determined by the plot in velocity is specific acceleration during that

specific time. The acceleration can vary during each time interval with little differences,

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or even very large differences. Even in uniformly accelerated motion, the acceleration

during each time interval can be different due to sources of error. The average

acceleration from the plot of the acceleration is the average of all the acceleration points

plotted onto the graph.

7.

Calculate the initial speed of the cart using kinematic equations with any of themeasured data. Please show your calculation.Given: A=9.8sin t=1.3s d=0 Required: Vi

=9.8sin7.5

=1.28 m/s

Analysis: d=vi(t)+1/2a(t)

Solution : 0=Vi(1.3)+1/2(1.28)(1.3)

0=Vi(1.3)+1.0816-1.0816=1.3Vi

-0.832m/s=Vi

Conclusion:

In the end, the results of the lab were based on the relationship between position, velocityand acceleration for motion in a straight line. The value of was measured before starting the

experiment. was measured to be 7.5; therefore making the accepted value of acceleration to be

a=gsin ; a=9.8sin7.5; a=1.28 m/s.During the conduction of the lab, the results of the position

time graph was a positive parabola. The reason this occurred is due to the relationship of theposition and the time. While going up the ramp, the cart will pass points along the ramp, and

while going back down, the cart will pass these same points again. However, going up the ramp,

the distance away from the sensor is getting smaller, eventually until it reaches its potential, getgoes down the ramp, which makes the cart further away again; creating a positive parabola. With

a parabola, the distance between each point is getting closer and closer together while going up

the ramp, but getting further and further while going down the ramp; this is due to the velocity.This made my hypothesis correct because the graph has a positive parabola with the repetition of

a position while going up, and coming back down the ramp. The velocity time graph was a

positive slope starting at a point with a negative y-axis, but moving to positive y-axis. The reason

this occurred is due to the relations to the position time graph. In this scenario, the positive is[downwards]; making the velocity increasing in a positive manner. This means that velocity is

pulling the cart down as soon as it begins. This also makes sense because the ratio between the

position and time is increasing a constant number, meaning a positive increase in velocity as

well. Also, while going up the ramp, the slope of the position time graph is negative, explainingthe negative velocity in the beginning; a positive slope while going down, explaining the positive

velocity during the second time period. This made my hypothesis correct because the graph had a

positive linear slope which started negative and the increase in velocity shows the acceleration.The acceleration time graph was a straight horizontal line with a slope of zero at a positive y-

axis. Again, the positive in this scenario is [downward] meaning that the acceleration is always

downward. The acceleration does not increase or decrease because the acceleration shows theconstant increase of the velocity. The velocity increases by the same amount every time for ever

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time interval, making the acceleration a constant number. This made my hypothesis correct

because the graph had a horizontal straight line with a slope of zero with no increase in

acceleration because then it would not be considered uniformly accelerated motion. Going back

to accepted value of acceleration, which was a=1.28 m/s, however the average accelerationtaken from the graph is a=1.1m/s; this creates a 13% error which can be due the sources of

errors.Sources of Error:

Measurement of the angle the incline is not exact when measured with a protractor,therefore making the acceleration not exact

The cart being pushed is not pushed in an exact straight line therefore making theposition versus time graph also not exact. If the cart is being pushed a slight angle, it is

travelling more distance

The push of the cart will not end exactly 15cm before the sensor which is should,therefore not giving the most accurate results

Initial force from the cart push is not 100% accurate, which is the reason velocity versustime graph is not a straight linear line, which also results for the reason that accelerationchanges.

During the starting and ending point of the sensor, the data is taken away because it is notrelevant. However, the actual start and end point are arbitrarily assumed. Therefore

making the initial position not being the same as the final position.