Answer
(a) By using the graph of $x$ vs $t^2$.
(b) See the graph below
The equation that best fits the graph is roughly, $x=3.75t^2$
(c) $a_{x}=7.50\frac{m}{s^2}$
(d) The speed of the glider when it reaches the bottom of the track is $4.50\frac{m}{s}$.
Work Step by Step
We take the (+)ve x-direction to be along the surface of the track pointing downwards and also taking the initial position of the glider to be the origin of our system.
Since the motion of the glider is along a frictionless air track with constant acceleration, we can apply the kinematics equation for constant acceleration.
(a) We can re-graph the data by plotting $x$ vs $t^2$ graph, so that the data points fall close to a straight line.
For constant acceleration of the glider starting from rest,we have
$x=\frac{1}{2}a_{x}t^2$->(1)
So, the graph of $x$ vs $t^2$ is a straight line with slope $\frac{1}{2}a_{x}$.
(b) We can construct the graph of $x$ vs $t^2$ by using the readings from the graph given in the text.
Using a graphing calculater, we can roughly get the following graph.
We can now obtain the equation for the straight line that is the best fit to the data points of our graph.
Slope of the line,$m=\frac{3.00-1.50}{0.8-0.4}=3.75\frac{m}{s^2}$
Now,$x-3.00=3.75(t^2-0.8)$
$=>x=3.75t^2$
(c)From equation(1) the slope of the graph of $x$ vs $t^2$ is$\frac{a_{x}}{2}$. So
$a_{x}=2\times3.75=7.50\frac{m}{s^2}$
(d) Let us calculate the speed of the glider when it reaches the bottom of the track (which was released from rest at a distance of $1.35m$ from the bottom of the track).
$2a_{x}(x_{f}-x_{i})=v_{f}^2-v_{i}^2$
$v_{f}=\sqrt (2\times7.50\times1.35)=4.50\frac{m}{s}$
The speed of the glider when it reaches the bottom of the track is $4.50\frac{m}{s}$.
