Option D) 1.5 m/s² is the correct answer.
The magnitude of the acceleration of the car is 1.5 m/s².
given the data in the question;
record of the observation;
Time(s) Speed(m/s)
3.0 4.0
5.0 7.0
6.0 8.5
Since its undergoing constant acceleration, we can make use any two from the three data recorded.
So, lets consider motion between t₁ = 3.0s and t₂ = 5.0s
Initial velocity at t₁ ; u = 4.0 m/s
final velocity at t₂; v = 7.0 m/s
Now, Time Elapsed; [tex]t = t_2 - t_1 = 7.0s - 5.0s = 2.0s[/tex]
To determine the magnitude of acceleration "a", we use the first equation of motion which says;
[tex]v = u + at[/tex]
we make "a" the subject of the formula
[tex]a =\frac{v - u}{t}[/tex]
so we substitute in our values;
[tex]a = \frac{(7.0m/s) - (4.0m/s)}{2.0s}[/tex]
[tex]a = \frac{3.0m/s}{2.0s}[/tex]
[tex]a = 1.5 m/s^2[/tex]
Therefore, the magnitude of the acceleration of the car is 1.5 m/s².
Option D) 1.5 m/s² is the correct answer.
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