Variable Acceleration

Section 4.8 Variable Acceleration

In the last section we studied motion in an interval in which acceleration is constant. That happens when force on the body is unchanging. However, many interesting situations, if not most, involve changing acceleration. For instance, when a ball hits the floor, force immediately changes from simple gravity to a combination of gravity and force from the floor. In this case, acceleration which has a value of \(g\) pointed down an instant before touching the ground but it changes to something else right after the ball comes in contact with the ball.

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To illustrate the mathematical process for dealing with a variable acceleration, it is sufficient to consider one-dimensional motion along \(x\)-axis. Then, we will have only \(a_x\) to deal with. For simplicity, we will look at \(a_x(t)\) that is made up of constant segments in the interval of interest. In this way, we can build full solution from constant-acceleration formulas applied to each infintesimal time-segments.

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Figure 4.73 shows acceleration in a one-dimensional situation along \(x\)-axis that has two values of acceleration in the interal \((0, t_2)\) of interest. Suppose \(v_{ix}=v_0\) and \(x_i=0\) at \(t=0\text{.}\) We wish to know \(v_x\) and \(x\) at \(t=t_2\text{.}\)

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In this situation, we first work out \(v_x\) and \(x\) at \(t=t_1\) and use that to work on the second segment. Let us denote them by \(v_1\) and \(x_1\text{.}\)

\begin{equation*} v_1 = v_0 + a_1 t_1;\ \ x_1 = v_0t_1 + \frac{1}{2}a_1t_1^2. \end{equation*}

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Now, we use these values as initial values for the second segment to obtian \(v_2\) and \(x_2\) at \(t=t_2\) using the acceleration of the second segment. Note the time during this segment is not \(t_2\) but \(t_2-t_1\text{.}\)

\begin{align*} v_2 \amp = v_1 + a_2 (t_2-t_1) = v_0 + a_1 t_1 + a_2 (t_2-t_1)\\ x_2 \amp = x_1 + v_1(t_2-t_1) + \frac{1}{2}a_2 (t_2-t_1)^2 \\ \amp = v_0t_1 + \frac{1}{2}a_1t_1^2 + ( v_0 + a_1 t_1 )(t_2-t_1) + \frac{1}{2}a_2 (t_2-t_1)^2 \end{align*}

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Subsection 4.8.1 (Calculus) Arbitrarily Varying Acceleration

Although most of our examples in this book will have constant-acceleration segments, it is worthwhile to look at a case of arbitrarily varying acceleration as an important extension of the multi-step method presented above. We will assume only \(a_x\) is nonzero, and, for brevity, use \(a\) for \(a_x\text{.}\)

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The basic idea for handling an arbitrarily varying acceleration is to replace the original acceleration by an approximation obtained by dividing up the time interval of interest into smaller time intervals. If time segments are small enough, we can approximate the original arbitrarily varying acceleration by a step-wise varying acceleration, where each step is an average acceleration in the corresponding time segment. This is often called discretizing acceleration.

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Clearly the original situation of a continuously varying acceleration is not the same as its replacement by constant acceleration steps. However, Isaac Newton showed that if intervals were allowed to be arbitrarily small, then the predictions of final position and velocity, based on step-wise approximate acceleration, can be made arbitrarily close to the exact answer without the approximation (see Figure 4.74.)

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Figure 4.74. A varying acceleration (a) and successive approximate constant acceleration steps of different step sizes (b, c and d). In (b) the variable acceleration is replaced by one constant step with the average acceleration denoted as \(a_1\) in (b). In (c) the variable acceleration is replaced by two segments of constant acceleration steps of \(a_1\) and \(a_2\text{.}\) Similarly, in (d) the variable acceleration is replaced by four steps. The process can be continued ad-infinitum. As you decrease the step size, or equivalently increase the number of steps, the original acceleration is covered more accurately by the approximation.
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With this approximation scheme, it is possible to find changes in velocity over the entire finite interval, from \(t = 0\) to \(t = T\text{,}\) not just over one of the small steps.

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Let us divide the interval \([0,T]\) into \(N\) equal intervals: \([0, \Delta t)\text{,}\) \([\Delta t, 2\Delta t )\text{,}\) \([2\Delta t, 3\Delta t)\text{,}\) \(\cdots\) \([(N-1)\Delta t, N\Delta t]\text{.}\) The time \(N\Delta t\) is equal to the total time \(T\) of all the intervals added. Let acceleration in thes subintervals be \(a_1\text{,}\) \(a_2\text{,}\) \(a_3\text{,}\) \(\cdots\) , \(a_N\text{,}\) respectively. Let \(v_1\text{,}\) \(v_2\text{,}\) \(v_3\text{,}\) \(\cdots\) , \(v_N\) be velocities at the end of the corresponding intervals. Of course \(v_N\) is same as \(v(T)\) or simply \(v\) at the end of the final time. Now, we show that our procedure leads us to the prediction of the final velocity from the initial velocity \(v_0\text{,}\) the velocity at time \(t=0\text{.}\)

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Interval	Velocity at the end of the interval
\(0 \text{ to } \Delta t \)	\(v_1 = v_0 + a_1 \Delta t\)
\(\Delta t \text{ to } 2\Delta t \)	\(v_2 = v_1 + a_2 \Delta t\)
\(2\Delta t \text{ to } 3\Delta t \)	\(v_3 = v_2 + a_3 \Delta t\)
\(3\Delta t \text{ to } 4\Delta t \)	\(v_4 = v_3 + a_4 \Delta t\)
\(\vdots \)	\(\vdots \)
\((N-2)\Delta t \text{ to } (N-1)\Delta t \)	\(v_{N-1} = v_{N-2} + a_{N-1} \Delta t\)
\((N-1)\Delta t \text{ to } N\Delta t \)	\(v_{N} = v_{N-1} + a_{N} \Delta t\)
\(0 \text{ to } N\Delta t \)	\(v_{N} = v_{0} + \sum_{i=1}^{N}a_{i} \Delta t\)

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Summing the velocity change equation in each interval gives us the following equation for the change in velocity from \(v_0\) to \(v\) over the entire time \([0,T]\text{.}\)

\begin{equation} v = v_0 + \sum_{i=1}^{N} a_i \Delta t. \tag{4.53} \end{equation}

The approximation becomes better as \(\Delta t\) is made smaller. We denote the limit of the summation as \(\Delta t\) approaches zero by another symbol, called the definite integral of \(a(t)\) from \(t = 0\) to \(t = T\text{.}\)

\begin{equation} v - v_0 = \int_{0}^{T} a(t) dt, \ or, (\text{Area under the curve of }a \text{ vs } t).\tag{4.54} \end{equation}

For an arbitrary interval t = t1 to t= t2, the integration will be done accordingly.

\begin{equation*} v_2 - v_1 = \int_{t_1}^{t_2} a(t) dt. \end{equation*}

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Similar arguments will lead you to analogous relation between change in position and velocity as we found between change in velocity and acceleration. We give the result for future reference and leave the derivation as an exercise for the student. Denoting average velocities in segments \(1, 2, 3, \cdots, N\) as \(\bar v_1, \bar v_2, \bar v_3, \cdots, \bar v_N\) we will obtain the following change in position \(x-x_0\) for the total time \(T=N\Delta t\text{.}\)

\begin{equation*} x - x_0 = \sum_{i=1}^N \bar v_i\Delta t \Longrightarrow \int_0^T v(t) dt,\ or, (\text{Area under the curve of }v \text{ vs } t). \end{equation*}

For an arbitrary interval t = t1 to t= t2, the integration will be done accordingly.

\begin{equation*} x_2 - x_1 = \int_{t_1}^{t_2} v(t) dt. \end{equation*}

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Example 4.75. Stepwise Varying Acceleration.

A train starts out at rest and moves in a straight line. The acceleration of the train changes with time, with magnitude \(0.5 \text{ m/s}^2\) between \(t = 0\) and \(t = 1 \text{ sec}\text{,}\) and then with magnitude \(1 \text{ m/s}^2\) between \(t = 1 \text{ sec}\) and \(t =3\ \text{sec}\) as shown in Figure 4.76. Find the position and the velocity of the train at \(t = 3 \text{ sec}\text{.}\)

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Answer.

\(2.5\text{ m/s}\)

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Solution.

As discussed above, we work out each constant acceleration segment one after the other.

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Interval \(t = 0\) to \(t =1 \text{ sec}\text{:}\)

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Let’s list the information about this interval as follows.

\begin{align*} \amp x_i = 0\ \text{(setting the initial position at the origin);}\\ \amp v_{ix} = 0\ \text{ (initially at rest);}\\ \amp a_x = 0.5 \text{ m/s}^2\ \text{ (constant -acceleration during this interval);}\\ \amp \text{Set t = 0 at the beginning of the interval so that we can use the standard formulas.} \\ \amp t = 1 \text{ s} \ \text{ (duration of this interval);} \end{align*}

With this information, we need to find \(x\) and \(v\text{,}\) the position and velocity at the end of interval.

\begin{align*} \amp v_x = v_{ix} + a_x t = 0 + 0.5\times 1 = 0.5 \text{ m/s};\\ \amp x = x_i + v_{ix} t + \frac{1}{2}a_xt^2 \\ \amp = 0 + 0 + \frac{1}{2}\times 0.5\times 1^2 = 0.25 \text{ m}. \end{align*}

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Interval \(t = 1 \text{ sec}\) to \(t =3 \text{ sec}\text{:}\)

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Using the information at the end of the previous segment of time, we now know the following information for this interval.

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\(x_i = 0.25 \text{ m}\) since initial \(x\)-position here is the final \(x\)-position at the end of the previous interval.

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\(v_{ix} = 0.5 \text{ m/s}\) since initial \(x\)-velocity here is the final \(x\)-velocity at the end of the previous interval.

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\(a_x = 1.0 \text{ m/s}^2\) since constant \(x\)-acceleration during this interval.

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\(t = 2 \text{ s}\) duration of this interval is from \(t= 1\text{ s}\) to \(t = 3 \text{ s}\) or \((3\text{ s}-1\text{ s}=2\text{ s})\text{.}\)

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With this information, we need to find \(x\) and \(v_x\text{,}\) the \(x\)-position and \(x\)-velocity at the end of interval.

\begin{align*} \amp v_x = v_{ix} + a t = 0.5 + 1\times 2 = 2.5 \text{ m/s};\\ \amp x = x_i + v_{ix} t + \frac{1}{2}a_xt^2 \\ \amp = 0 + 0.5\times 2 + \frac{1}{2}\times 1.0\times 2^2 = 3.0 \text{ m}. \end{align*}

Therefore, the train is at \(3.0 \text{ m}\) from where it was at \(t=0\) and it is moving at \(2.5\text{ m/s}\) at \(t=3.0 \text{ s}\) mark.

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Example 4.77. Two Acceleration Segments.

A particle starts out at rest at \(t=0\) and accelerates in a straight line with an acceleration of \(2 \text{ m/s}^2\) from \(t = 0\) to \(t = 3\text{ sec}\text{,}\) and then with an acceleration of \(4\ \text{m/s}^2\) pointed in the same direction from \(t = 3\text{ sec}\) to \(t =5\text{ sec}\text{.}\)

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(a) Find the position and the velocity of the particle at \(t = 3\) sec.

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(b) Find the position and the velocity of the particle at \(t = 5\) sec.

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Answer.

(a) \(9\text{ m},\ 6\text{ m/s}\)m (b) \(29\text{ m},\ 14\text{ m/s}\text{.}\)

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Solution 1. a

Working between \(t=0\) and \(t=3\text{ sec}\) we get

\begin{align*} v_f \amp = v_i + a t = 0 +2\text{ m/s}^2\times 3\text{ s} = 6\text{ m/s}. \\ x_f \amp = x_i + \dfrac{v_i + v_f}{2}\, t = 0 +3\text{ m/s}\times 3\text{ s} = 9\text{ m}. \end{align*}

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Solution 2. b

When, we work between \(t=3\text{ sec}\) and \(t=5\text{ sec}\text{,}\) the initial condition is same as final condition in the last step. Therefore we jnow have

\begin{equation*} x_i = 9\text{ m},\ \ v_i = 6\text{ m/s}. \end{equation*}

With \(a=4\text{ m/s}^2\text{,}\) we get

\begin{align*} v_f \amp = v_i + a t = 6\text{ m/s} +4\text{ m/s}^2\times 2\text{ s} = 14\text{ m/s}. \\ x_f \amp = x_i + \dfrac{v_i + v_f}{2}\, t = 9\text{ m} +\dfrac{6\text{ m/s} + 14\text{ m/s}}{2}\times 2\text{ s} = 29\text{ m}. \end{align*}

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Example 4.78. Variable Acceleration - Two Acceleration Segments.

At \(t=0\) a particle located at origin has a velocity of \(20\) m/s in the direction of the positive \(x\)-axis, and accelerates on the \(x\)-axis as shown in the figure. The \(y\) and \(z\)-components of the acceleration are zero.

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(a) Find the position and the velocity of the particle at \(t = 2\) sec.

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(b) Find the position and the velocity of the particle at \(t = 5\) sec.

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Answer.

(a) \(34\text{ m},\ 14\text{ m/s}\)m (b) \(94\text{ m},\ 26\text{ m/s}\text{.}\)

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Solution 1. a

Working between \(t=0\) and \(t=2\text{ sec}\) we have

\begin{equation*} x_i=0,\ v_i=20\text{ m/s},\ a = -3\text{ m/s}^2,\ t = 2\text{ s}. \end{equation*}

Therefore,

\begin{align*} v_f \amp = v_i + a t = 20\text{ m/s} + (-3\text{ m/s}^2)\times 2\text{ s} = 14\text{ m/s}. \\ x_f \amp = x_i + \dfrac{v_i + v_f}{2}\, t = 0 +17\text{ m/s}\times 2\text{ s} = 34\text{ m}. \end{align*}

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Solution 2. b

When, we work between \(t = 2\text{ sec}\) and \(t = 5\text{ sec}\text{,}\) the initial condition is same as final condition in the last step. Therefore we now have

\begin{equation*} x_i = 34\text{ m},\ v_i = 14\text{ m/s},\ a = 4\text{ m/s}^2,\ t = 3\text{ s}. \end{equation*}

Therefore,

\begin{align*} v_f \amp = v_i + a t = 14\text{ m/s} + 4\text{ m/s}^2\times 3\text{ s} = 26\text{ m/s}. \\ x_f \amp = x_i + \dfrac{v_i + v_f}{2}\, t = 34\text{ m} +\dfrac{14\text{ m/s} + 26\text{ m/s}}{2}\times 3\text{ s} = 94\text{ m}. \end{align*}

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Exercises 4.8.2 Exercises

1. Particle with Three Acceleration Segments.

At \(t=0\) a particle located at origin has a velocity of \(20\) m/s in the direction of the positive \(x\)-axis, and accelerates on the \(x\)-axis as shown in Figure 4.79>. The \(y\) and \(z\)-components of the acceleration are zero.

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(a) Find the position and the velocity of the particle at \(t = 2\) sec.

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(b) Find the position and the velocity of the particle at \(t = 3\) sec.

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Answer.

(a) \(30\,\text{m}\text{,}\) \(10\,\text{m/s}\text{;}\) (b) \(40\,\text{m}\text{,}\) \(10\,\text{m/s}\text{;}\) (c) \(70\,\text{m}\text{,}\) \(20\,\text{m/s}\text{;}\)

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Solution 1. a

For all three, we will use the following technique. Denote the interval by \(t_0\le t \le t_f\text{,}\) duration \(T = t_f - t_0\text{,}\) initial position by \(x_0\text{,}\) final position by \(x\text{,}\) initial velocity by \(v_0\text{,}\) final velocity by \(v\text{,}\) and acceleration by \(a\text{.}\) For brevity we are dropping the subscript \(x\) from the symbols. The formulas we need are:

\begin{align*} \amp v = v_0 + a T \\ \amp x = x_0 + v_0 T + \frac{1}{2} T^2 \end{align*}

Note the time in these formulas are the interval and the actual instant times. Also, although \(x_0\) will be zero for the first interval, it will be nonzero for other intervals.

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Here (drop units, for now)

\begin{equation*} t_0 = 0,\ t_f = 2,\ x_0=0,\ x = ?,\ v_0 = 20,\ v = ?,\ a = -5. \end{equation*}

Hence, with \(T=t_f-t_0 = 2\) we get

\begin{align*} \amp v = 20 + (-5) 2 = 10\,\text{m/s}.\\ \amp x = 0 + 20\times 2 + \frac{1}{2}\times (-5) \times 2^2 = 30\text{ m}. \end{align*}

These will be the intial values for the next time segment!

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Solution 2. b

Here (drop units, for now)

\begin{equation*} t_0 = 2,\ t_f = 3,\ x_0=30,\ x = ?,\ v_0 = 10,\ v = ?,\ a = 0. \end{equation*}

Hence

\begin{align*} \amp v = 10 + 0 = 10\,\text{m/s}.\\ \amp x = 30 + 10\times 1 + 0 = 40\text{ m}. \end{align*}

These will be the intial values for the next time segment!

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Solution 3. c

Here (drop units, for now)

\begin{equation*} t_0 = 3,\ t_f = 5,\ x_0=40,\ x = ?,\ v_0 = 10,\ v = ?,\ a = 5. \end{equation*}

Hence

\begin{align*} \amp v = 10 + 5\times 2 = 20\,\text{m/s}.\\ \amp x = 40 + 10\times 2 + \frac{1}{2}\times 5 \times 2^2 = 70\text{ m}. \end{align*}

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2. Box Sliding with Linear Time Acceleration Segments.

A box slides on a floor in a straight path such that its acceleration is not constant in time, but given by Figure 4.80. (a) Find the velocity of the box at t = 4 sec if it starts out at rest at t = 0. (b) Find the velocity of the box at t = 10 sec.

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Answer.

(a) \(6\text{ m/s}\text{,}\) (b) \(15\text{ m/s}\text{.}\)

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Solution 1. a

(We will drop \(x\) from subscripts as well as units.) In any interval, \([t_i,\, t_f]\text{,}\) wfor each Cartesian component of velocity and acceleration, we have the following relation (from definition)

\begin{equation*} v_f - v_i = \int_{t_i}^{t_f}\, a(t)\, dt. \end{equation*}

From the given plot of \(a_x\) vs \(t\) we have the following for the time in the interval \([0,\, 4\,\text{s}]\text{:}\)

\begin{equation*} a = \frac{3}{4}\,t. \end{equation*}

Therefore, since \(v_0=0\) we have

\begin{equation*} v_4 - 0 = \int_0^4 \frac{3}{4}\,t\, dt = 6\text{ m/s}. \end{equation*}

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Solution 2. b

For this segment, the slope is \(-1/2\text{,}\) but it also need to go through \((10, 0)\text{.}\) Hence, we have

\begin{equation*} a - 0 = -\frac{1}{2}\,(t - 10)\ \rightarrow\ \ a = 5 - \frac{1}{2} t. \end{equation*}

From part (a), we have the velocity at the initial instant of this interval to be \(v_4 = 6\text{.}\) Hence

\begin{equation*} v_6 - 6 = \int_4^10 \left(5 - \frac{1}{2} t \right) dt =9. \end{equation*}

Hence, \(v_6 = 15\,\text{m/s}\text{.}\)

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3. Time Varying Deceleration of a Hockey Puck.

A hockey puck is shot on a surface that has different roughness at different places. As a result its acceleration (i.e. deceleration) varies from place to place, which can be plotted as a function of time starting from the instant puck leaves the stick. The puck has a velocity of \(2\,\text{m/s}\) at \(t = 5\,\text{sec}\text{.}\) Find the velocity of the puck at \(t = 0\,\text{sec}\text{.}\)

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Answer.

\(21.25\text{ m/s}\text{.}\)

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Solution.

We will implement the following equation.

\begin{equation*} v_f - v_i = \int_{t_i}^{t_f}\, a(t)\, dt. \end{equation*}

Here \(a(t)\) is piecewise function:

\begin{equation*} a(t) = \begin{cases} -6 + t \amp 0 \le t \le 2 \\ -4 \amp 2 \le t \le 3.5 \\ -1 + 2 (t-5) \amp 3.5 \le t \le 5 \end{cases} \end{equation*}

Since \(v_5 = 2\text{ m/s}\text{,}\) the integration gives

\begin{equation*} 2 - v_0 = \int_0^2 (-6 + t) dt + (-4)\int_{2}^{3.5} dt + \int_{3.5}^{5} (-11 + 2t) dt = -19.25. \end{equation*}

Hence, \(v_0 = 21.25\text{ m/s}\text{.}\)

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4. (Calculus) Change in Position and Velocity from Analytic Expression of Variable Acceleration.

The \(x\)-component of the acceleration of a particle changes according to the following analytic expression, \(a_x(t) = 4+t^2/2\text{,}\) where \(t\) is in sec and the acceleration in \(\text{m/s}^2\text{.}\) Find the \(x\)-components of the velocity and position vectors. Use \(v_0\) and \(x_i\) for initial velocity and initial position respectively.

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Hint.

\(v_x(t) - v_x(0) = \int_0^t\, a_x dt\text{.}\)

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Answer.

(a) \(x = x_i + v_0 t + 2 t^2 + \dfrac{1}{24} t^4\text{,}\) (b) \(v = v_0 + 4 t + \dfrac{1}{6} t^3 \text{.}\)

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Solution.

Integrating \(a_x(t)\) will give \(v_x(t) - v_0\) and integrating \(v_x(t)\) will give \(x(t)-x_i\text{.}\)

\begin{align*} v_x(t) \amp = v_x(0) + \int_0^t\, a_x dt \\ \amp = v_0 + \int_0^t\,\left( 4+t^2/2 \right) dt \\ \amp = v_0 + 4 t + \dfrac{1}{6} t^3. \\ x(t) \amp = x(0) + \int_0^t\, v_x(t) dt \\ \amp = x_i + \int_0^t\, \left(v_0 + 4 t + \dfrac{1}{6} t^3 \right) dt \\ \amp = x_i + v_0 t + 2 t^2 + \dfrac{1}{24} t^4. \end{align*}

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5. Practice with a Friend: Accelerate and Decelerate.

Suppose you accelerate your car at a constant rate of \(2.0\text{ m/s}^2\) starting from rest for some unspecified time. After that you immediately apply your brake and decelerate at \(5.0\text{ m/s}^2\) and come to a stop. The total distance covered between start and stop was \(800\text{ m}\text{.}\) (a) Find the peak velocity. (b) Find the total time it was in the accelerating phase. (c) Find the total time it was in the decelerating phase.

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