Motion under Constant Acceleration

Recall that the position and the acceleration of an object are related to each other by the second derivative. If the position of an object is a function \(x(t)\), then the acceleration is

\[ a(t)=\dfrac{d^2}{dt^2}x(t).\]

Since the derivative of a function at a given point is the slope of the function at that point, the velocity of an object is the slope of the displacement function and the acceleration of an object is the slope of the velocity function.

When the acceleration of a function is constant, the slope of the velocity function is also constant, i.e.

\[a(t)=\dfrac{d}{dt}v = \text{constant}.\]

Since the velocity function is the derivative of the position function, the position function can be obtained from the acceleration by integrating twice:

\[x - x_0 = v_0 t + \frac12 at^2,\]

where \(x_0\) is the initial position and \(v_0\) is the initial velocity.

One can see that the graph of the position function over time will be a parabola, since the equation for \(x(t)\) above is quadratic in time \(t\). Taking the second derivative of the \(\frac12 at^2\) term yields the constant acceleration \(a\).

Plot the position-time graph of a particle undergoing constant acceleration with \(a = 5 \text{ m}/\text{s}^2\) from rest at \(x = 0\), and the position-graph of a particle undergoing constant acceleration with initial velocity \(v = 1 \text{ m}/\text{s}\) at \(x = 0\).

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Below the two graphs are plotted, with the particle starting at rest in blue and the particle starting with positive initial velocity in gold:

Position-time graphs of two particles undergoing constant acceleration.

Both position-time graphs are parabolic; the gold curve has a nonzero slope at \(x = 0\) corresponding to the positive initial velocity at \(x=0,\) while the blue curve is flat at \(x=0\) corresponding to starting at rest. The gold curve increases faster, because the initial velocity of the second particle causes the second particle to accumulate more distance in the same amount of time.

Recall that acceleration is defined as the time rate of change of the velocity of some mass, that is,

\[a = \frac{dv}{dt}.\]

Given a constant acceleration \(a\), the change in velocity \(\Delta v\) of this mass after a time \(t\) is therefore

\[\Delta v = a t.\]

Assuming that the mass starts at a velocity \(v_0\), the velocity over time of this mass is therefore

\[v(t) = v_0 + \Delta v = v_0 + at.\]

This says that the velocity over time is the original velocity plus however much one has accelerated in a given amount of time.

To derive the position over time given constant acceleration, one must now use calculus. Recall that the velocity is defined as the time rate of change of the position:

\[v = \frac{dx}{dt}.\]

The position over time is therefore found by integrating both sides (readers familiar with calculus should check this!). If the initial position of the mass is \(x_0\), the result is

\[x(t) = x_0 + v_0 t + \frac12 at^2.\]

A ball rolling down a hill constantly accelerates at \(0.6 \text{ m/s}^{2}\) from an initial velocity of \(3 \text{ m/s}\). If it takes \(10\text{ s}\) to reach the bottom, find the distance that the ball rolled.

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As acceleration is constant in this scenario, we can apply the formula \(x(t) = x_0 + v_0 t + \frac12 at^2\), which gives \[x = 0 + 3 \times 10 + \frac{1}{2} \times 0.6 \times 10^2 = 60 \text{ m}.\] Thus the ball rolled a total of \(60 \text{ m}\).