We can use the formula for uniform acceleration to solve the problem:

$a = \frac{v_f - v_i}{t_f - t_i}$

Where:

• $v_f$ is the final velocity,
• $v_i$ is the initial velocity,
• $t_f$ is the final time,
• $t_i$ is the initial time.

From the data, we can select two points. Let’s use the first and last points:

1. $v_i = 4.0 \, \text{m/s}$ at $t_i = 1.0 \, \text{s}$
2. $v_f = 19.0 \, \text{m/s}$ at $t_f = 7.0 \, \text{s}$

Now, substitute these values into the formula:

$a = \frac{19.0 \, \text{m/s} - 4.0 \, \text{m/s}}{7.0 \, \text{s} - 1.0 \, \text{s}} = \frac{15.0 \, \text{m/s}}{6.0 \, \text{s}} = 2.5 \, \text{m/s}^2$

Thus, the magnitude of the acceleration is 2.5 m/s².

Would you like more details on this solution, or do you have any questions?

Here are 5 questions related to this:

1. How do we determine uniform acceleration from a set of data points?
2. What if the car's acceleration was not uniform—how would the calculation change?
3. Can we use another pair of points to confirm the acceleration value?
4. How is acceleration related to the area under a velocity-time graph?
5. What are some common real-life examples of uniform acceleration?

Tip: Always choose data points far apart to minimize the effect of rounding errors when calculating acceleration.