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Four cars start from rest. Car A accelerates at 6.0 m/s', Car B accelerates at 5.4 m/s', Car C at 8.0 m/s' and Car D at 12.0 m/s'. (A) Fill in the velocity and displacement columns for

each car at t = 2.0 s in the table below. (B) What conclusion do you reach about the
velocity attained and the displacement of a body starting from rest at the end of the first 2.0 s of acceleration?

1 Answer

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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

---

V = vi + (2)*a

V = (0) + (2)*a

V = 2*a

For vi = 0 and t = 2, D = V!

Four cars start from rest. Car A accelerates at 6.0 m/s', Car B accelerates at 5.4 m-example-1
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