Answer
a) Yes.
b) $7.86\;\rm m/s^2$
Work Step by Step
b)
First of all, we need to use the kinematic formula of velocity squared.
$$v_{fx}^2-v^2_{ix}=2a_x\Delta x$$
We know that the final velocity of the car is always zero.
$$0-v^2_{ix}=2a_x(x-x_i)$$
Let's assume that the initial position is always at $x=0$,
$$-v^2_{ x}=2a_xx$$
we will ignore the negative sign since we are seeking the magnitude of the deceleration of the car.
$$ v^2_{x}=2a_x x$$
Now we can draw the velocity squared as a function of position by plugging the data on the given table.
Recall the equation of the linear line $y=mx+b$
whereas $m$ is the slope of the line and $b$ is the constant that intercepts the $y$-axis. And in our case, $y=v_x^2$, $m=2a_x$, $b=0$, and $x=x$.
After we plugged the points, we need to draw the best fit kine, as you see in the figure below.
Now we can find the slope of the best fit line to find the deceleration of the car.
From the graph below;
$${\rm Slope}=\dfrac{978.99-0}{62.28-0}=2a_x=15.72$$
Thus,
$$a_x=\color{red}{\bf 7.86}\;\rm m/s^2$$
a) Therefore, yes, the data support an assertion that the deceleration is constant, independent of speed.
