Question: In your physics lab you release a small glider from rest at various points on a long, frictionless air track that is inclined at an angle above the horizontal. With an electronic photocell, you measure the time it takes the glider to slide a distance from the release point to the bottom of the track. Your measurements are given in Fig. P2.84, which shows a second-order polynomial (quadratic) fit to the plotted data. You are asked to find the glider’s acceleration, which is assumed to be constant. There is some error in each measurement, so instead of using a single set of x and t values, you can be more accurate if you use graphical methods and obtain your measured value of the acceleration from the graph. (a) How can you re-graph the data so that the data points fall close to a straight line? (Hint: You might want to plot or , or both, raised to some power.) (b) Construct the graph you described in part (a) and find the equation for the straight line that is the best fit to the data points. (c) Use the straight line fit from part (b) to calculate the acceleration of the glider. (d) The glider is released at a distance x = 1.35 from the bottom of the track. Use the acceleration value you obtained in part (c) to calculate the speed of the glider when it reaches the bottom of the track.

The given data is listed below as:

• The angle of inclination of the glider is $\theta$,
• The distance from the release point to the track’s bottom point is, x
• The time taken by the glider to reach the distance is, t
• The distance of the glider from the track’s bottom is, x= 1.35m

The acceleration of an object is described as the rate at which the velocity changes with the change in time. The acceleration is the division of the rate of change of velocity with time.

For making the data points fall along a straight line, a$x-t^2$ graph is required. The reason that the graph is required is that with the increase of the time, the distance also increases but if the time increases at a fast face, the distance will also increase then the $x-t^2$graph will be a straight line.

Thus, a$x-t^2$ graph is required to make the data points fall close to a straight line.

The graph$x-t^2$ has been provided below:

Here, from the above graph, it has been identified that the data points fall along a straight line. Moreover, the equation of the straight line has also been obtained with the help of the data points that is$x=3.8t^2+0.009$ .

The equation obtained from the graph is expressed as:

$x=3.8t^2+0.009$ …(i)

Here, x is the distance from the release point to the track’s bottom point and t is the time taken by the glider to reach the distance.

Thus, the equation of the straight line that best fits to the data points is $x=3.8t^2+0.009$.

The equation obtained from the graph is expressed as:

$x=3.8t^2+0.009$

Here, x is the distance from the release point to the track’s bottom point and t is the time taken by the glider to reach the distance.

Double differentiating the above equation with respect to the time , the above equation can be expressed as:

$$\begin{gathered} \frac{dx}{dt}=7.6t \\ =v \\ \frac{d^{2}x}{d{t}^2}=7.6 \\ =a \end{gathered}$$ …(ii)

As the acceleration is described as the rate of the change of velocity with time, then the acceleration of the glider is $7.6{\text{m/s}}^{\text{2}}$.

Thus, the acceleration of the glider is $7.6{\text{m/s}}^{\text{2}}$.

Recalling the equation (i) in order to find the speed of the glider.

$x=3.8t^2+0.009$

Here, x is the distance from the release point to the track’s bottom point and t is the time taken by the glider to reach the distance.

Substitute x fori.35 in the above equation.

$$\begin{gathered} 1.35=3.8t^2+0.009 \\ 3.8t^2=1.341 \\ {\text{t}}^2=0.352 \\ t=0.593 \end{gathered}$$

The negative value t of the time is neglected as the time cannot be negative.

Here, the value of the time t has been obtained as 0.53s. With the help of this, the speed of the glider can be obtained.

From the equation (ii), the speed of the glider can be expressed as:

v = 7.6 t

Substitute for in the above equation.

$$\begin{gathered} v=7.6\times \left(0.593\right) \\ =4.509 \end{gathered}$$

Hence, the speed of the glider when it reaches the bottom of the track is$4.509\text{m/s}$ .

Thus, the speed of the glider when it reaches the bottom of the track is$4.509\text{m/s}$ .