Answer:
Under the conditions set in this problem, the numerical values for distance (m) and velocity (m/2) for the 4 cars are, surprisingly, identical. A derivation of why this is true is provided in the explanation. The results are attached.
Step-by-step explanation:
The equation for distance calculated from acceleration is
D = vi + 1/2a*t^2,
where v is the velocity, t is time, and a is the acceleration. vi is the initial velocity (t = 0) when the acceleration, a, begins.
The velocity of an object undergoing acceleration is given by:
V = vi + t*a, where vi is the initial velocity, if any. a may be expressed in units of distance (e.g., meters) per time squared. A common unit is m/s^2 (meters per second per second).
A spreadsheet with the cars and their accelerations is attached. The formulas above were used to generate the results for Distance (D, in meters) and Velocity (V, in m/s).
An unexpected result for this particular set of conditions is that the values for both distance travelled and velocity after 2 seconds are numerically the same.
We can see why this is if we compare the two equations for Distance and Velocity as a function of time, t.
D = vi + 1/2a*t^2
V = vi + t*a
Since we were give time of 2 seconds and an initial velocity of 0, these equations become:
D = vi + 1/2a*(2)^2
D = (0) + (1/2)*(4)*a
D = 2*a
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V = vi + (2)*a
V = (0) + (2)*a
V = 2*a
For vi = 0 and t = 2, D = V!