Car data!

(a) Remember what it means for the motion of an object to have a positive, negative or zero acceleration. Think about the implications of an acceleration that is proportional to time.

(b) Use the equation of motion for an object moving with constant acceleration. Treat each of the four regions independently, and remember to find the velocity in each region.

(c) All the standard kinematic equation are for objects moving with constant acceleration. The equations will be very different if the acceleration is linear!

(d) Use the values found in (a) for the velocity.

(e) Use the values found in (a) for the position.

(a) From 10 to 20 seconds, the acceleration is constant and negative. Then, the rate of change of the velocity over time is constant and negative. This indicates that the driver stopped pressing the throttle and started braking.

From 20 to 30 seconds, we have a positive acceleration, so the velocity is increasing. The driver is pressing the throttle to increase speed.

From 30 to 40 seconds, the acceleration is zero, which means the velocity is constant. The driver is neither braking nor pressing the throttle.

Finally, from 40 to 50 seconds, the acceleration is increasing in a constant manner. The velocity changes with time and is not only positive but the rate itself is increasing! So the velocity is increasing, and as time passes, it increases at an even faster rate. This is equivalent to the driver pressing the throttle a little harder each time, to change not only the velocity but the acceleration as well.

(b) To calculate the total displacement as:

\begin{equation*}
\vec{D}=\vec{x}(40\, \text{s})-\vec{x}(0\, \text{s}).
\end{equation*}

For all the regions, the general equation of motion will be useful:

\begin{equation*}
\vec{x}_f=x_{i}\,\hat{\textbf{i}}+v_{i}t\,\hat{\textbf{i}}+\frac{1}{2}at^2\,\hat{\textbf{i}},
\end{equation*}

and also the equation for the velocity:

\begin{equation*}
\vec{v}_f=v_{i}\,\hat{\textbf{i}}+at\,\hat{\textbf{i}},
\end{equation*}

First Region: \(0\leq t\leq 10\)

Using the numerical values, the position at \(t=10 \, \text{s} \) is:

\begin{equation*}
\vec{x}_1=200\,\text{m} \,\hat{\textbf{i}},
\end{equation*}

and the velocity is:

\begin{equation*}
\vec{v}_1=40\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation*}

Some values are useful to find because they will serve as the initial position and velocity for the next segment.

Second region: \(10\leq t\leq 20\)

Consider that we should write the time as \((t-t_2)\) instead of \(t\) in all equations here, where \(t_2 = 10 \, \text{s}\). For the position at time \(t=20\,\text{s}\), we get:

\begin{equation*}
\vec{x}_2=450\,\text{m}\,\hat{\textbf{i}}.
\end{equation*}

The velocity is:

\begin{equation*}
\vec{v}_2=10\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation*}

Third Region: \(20\leq t\leq 30\) 

Again, writing \((t-t_3)\) instead of \(t\) in all equations. For the position at time \(t=30\,\text{s}\), we get:

\begin{equation*}
\vec{x}_3=650\,\text{m}\,\hat{\textbf{i}}.
\end{equation*}

The velocity is:

\begin{equation*}
\vec{v}_3=30\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation*}

Fourth region: \(30\leq t\leq 40\)

In the fourth region, noting that it is a motion with constant velocity.  Again, we write \((t-t_4)\) instead of \(t\) for all equations. For the position at time \(t=40\,\text{s}\), we get:

The result is

\begin{equation*}
\vec{x}_4=950\,\text{m}\,\hat{\textbf{i}}.
\end{equation*}

The velocity is the same as before since is constant:

\begin{equation*}
\vec{v}_4=30\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation*}

Thus, the total displacement between 0 and 40 seconds is:

\begin{equation*}
\vec{D}=950\,\text{m}\,\hat{\textbf{i}}-0\,\text{m}\,\hat{\textbf{i}}.
\end{equation*}

(c) The acceleration between 40 and 50 seconds increases linearly with time. It is not a constant, clearly. We cannot use the usual relations between the kinematic variables as we have done in the previous parts of the problem. We would have to integrate twice using the definition for the acceleration in terms of position, that is, we would have to integrate twice the following expression:

\begin{equation*}
\vec{a}=\frac{d^2x}{dt^2}\,\hat{\textbf{i}}.
\end{equation*}

(d) Using the results of point b), we already know an expression for the velocity \(v\) for each region in terms of the time and other known variables. We can the make the graph in the following figure:

e) We can then use the results of all the position equations in terms of time that we found in b). If we do so, we get the graph in the following figure:

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) We need to interpret each segment of the given plot. Since we are looking at an acceleration graph, we know that we are looking at the rate at which the velocity of the car changes with time (recall that acceleration is change in velocity over time).

Keeping this in mind, we see that from 0 to 10 seconds the acceleration is constant and positive, meaning that the rate of change of the velocity over time is constant and positive. Hence,  the velocity increases during this interval. The driver is pressing the throttle to increase the speed (since the car is initially moving in the positive direction of X, then an increase of velocity is always an increase of speed, but this would not be the case if the car was moving in the opposite direction).

From 10 to 20 seconds, the acceleration is constant and negative, meaning that the rate of change of the velocity over time is constant and negative. That is, the velocity decreases. Physically, this indicates that the driver stopped pressing the throttle and started braking.

From 20 to 30 seconds, we have a positive acceleration again, and so, again, the velocity is increasing. The driver is pressing the throttle to increase speed.

From 30 to 40, seconds the acceleration is zero, which means the velocity is constant. The driver is neither braking nor pressing the throttle.

Finally, from 40 to 50 seconds, the acceleration is increasing in a constant manner. So the rate of change at which the velocity changes with time is not only positive but the rate itself is increasing! So the velocity is increasing, and as time passes, it increases at an even faster rate. This is equivalent to the driver pressing the throttle a little harder each time, to change not only the velocity but the acceleration as well.

b) To calculate the total displacement between 0 and 40 seconds, we will have to find the change in the position vector between these two times. In particular, the displacement in question is given by the difference in the final and initial positions:

\begin{equation}
\label{Displace}
\vec{D}=\vec{x}(40\, \text{s})-\vec{x}(0\, \text{s}).
\end{equation}

If we assume the car starts on the origin, then all we need to find is the position at 40 seconds, and to do that we need to consider the equation of motion for each segment until reaching that time. For this purpose, we will use that the velocity at time \(t=0\) is zero, that the car starts moving along the positive X axis, and that the car always moves in a straight line, as they say in the prompt.

First Region: \(0\leq t\leq 10\)

For this region, we write the equation of the position of an object moving with constant acceleration:

\begin{equation}
\vec{x}_1=x_{i,1}\,\hat{\textbf{i}}+v_{i,1}t\,\hat{\textbf{i}}+\frac{1}{2}a_1t^2\,\hat{\textbf{i}},
\end{equation}

where \(x_{i,1}=0\,\text{m}\) is the position at time \(t=0\,\text{s}\), \(v_{i,1}=0\,\text{m/s}\) is the speed at time \(t=0\,\text{s}\) and \(a_1=4\,\text{m/s}^2\) (and positive). Using this information, we get the following for the position at time \(t=10\, \text{s}\):

\begin{equation}
\vec{x}_1=\frac{1}{2}(4\,\text{m/s}^2)(10\,\text{s})^2 \,\hat{\textbf{i}},
\end{equation}

which yields

\begin{equation}
\label{x1i}
\vec{x}_1=200\,\text{m} \,\hat{\textbf{i}}.
\end{equation}

The velocity at time \(t=10\,\text{s}\) is calculated using the expression of the velocity in an accelerated motion, namely

\begin{equation}
\vec{v}_1=v_{i,1}\,\hat{\textbf{i}}+a_1t\,\hat{\textbf{i}},
\end{equation}

where we used the fact that, initially, the car moves in the positive direction of X. Using the numerical values, we get

\begin{equation}
\vec{v}_1=0\text{m/s}\,\hat{\textbf{i}}+(4\,\text{m/s}^2)(10\,\text{s})\,\hat{\textbf{i}},
\end{equation}

that is,

\begin{equation}
\label{v1i}
\vec{v}_1=40\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation}

The vectors given by equations \eqref{x1i} and \eqref{v1i} will be useful when writing the position and acceleration for the second region, since they will serve as the initial position and velocity for that segment.

Second region: \(10\leq t\leq 20\)

We can now write the equation for the position in a constant accelerating motion in the second region, taking into account that a time \(t_2=10\,\text{s}\) has already passed. So we should write \((t-t_2)\) instead of \(t\) in all equations here. Explicitly, we have

\begin{equation}
\vec{x}_2=x_{i,2}\,\hat{\textbf{i}}+v_{i,2}(t-t_2)\,\hat{\textbf{i}}+\frac{1}{2}a_2(t-t_2)^2\,\hat{\textbf{i}},
\label{x2}
\end{equation}

where \(x_{i,2}=200\,\text{m}\,\hat{\textbf{i}}\) is the position at time \(t=10\,\text{s}\) (found in the first segment), \(v_{i,2}=40\,\text{m/s}\,\hat{\textbf{i}}\) is the velocity at time \(t=10\,\text{s}\) (also found earlier) and \(a_2=-3\,\text{m/s}^2\,\hat{\textbf{i}}\) because the car has negative acceleration.

The position at time \(t=20\) is then, using the numerical values and equation \eqref{x2}, the following:

\begin{equation}
\vec{x}_2=200\,\text{m}\,\hat{\textbf{i}}+(40\,\text{m/s})(20\,\text{s}-10\,\text{s})\,\hat{\textbf{i}}+\frac{1}{2}(-3\,\text{m/s}^2)(20\,\text{s}-10\,\text{s})^2\,\hat{\textbf{i}},
\end{equation}

which results in

\begin{equation}
\label{x2i}
\vec{x}_2=450\,\text{m}\,\hat{\textbf{i}}.
\end{equation}

The velocity at time \(t=20\,\text{s}\) is calculated using the expression

\begin{equation}
\vec{v}_2=v_{i,2}\,\hat{\textbf{i}}+a_2(t-t_2)\,\hat{\textbf{i}}.
\end{equation}

Using the numerical values, this becomes

\begin{equation}
\vec{v}_2=40\text{m/s}\,\hat{\textbf{i}}+(-3\,\text{m/s}^2)(20\,\text{s}-10\,\text{s})\,\hat{\textbf{i}},
\end{equation}

that is,

\begin{equation}
\label{v2i}
\vec{v}_2=10\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation}

The results given by equations \eqref{x2i} and \eqref{v2i} will be useful when writing the position and acceleration for the third region. They will serve as the initial position and velocity, respectively.

Third Region: \(20\leq t\leq 30\) 

We can now write the equation for the position in a constant accelerating motion, taking into account that a time of \(t_3=20\,\text{s}\) has already passed. This means that we should write \((t-t_3)\) instead of \(t\) in all equations. Explicitly,

\begin{equation}
\vec{x}_e=x_{i,3}\,\hat{\textbf{i}}+v_{i,3}(t-t_3)\,\hat{\textbf{i}}+\frac{1}{2}a_3(t-t_3)^2\,\hat{\textbf{i}},
\end{equation}

where \(\vec{x}_{i,3}=450\,\text{m}\) is the position at time \(t=20\,\text{s}\) (found for the second segment), \(\vec{v}_{i,3}=10\,\text{m/s}\) is the velocity at time \(t=20\,\text{s}\)  (found in the second segment) and \(a_3=2\,\text{m/s}^2\,\hat{\textbf{i}}\) (it is positive).

The position at time \(t=30\) is then, using the numerical values, the following:

\begin{equation}
\vec{x}_3=450\,\text{m}\,\hat{\textbf{i}}+(10\,\text{m/s})(30\,\text{s}-20\,\text{s})\,\hat{\textbf{i}}+\frac{1}{2}(2\,\text{m/s}^2)(30\,\text{s}-20\,\text{s})^2\,\hat{\textbf{i}},
\end{equation}

which yields

\begin{equation}
\label{x3i}
\vec{x}_3=650\,\text{m}\,\hat{\textbf{i}}.
\end{equation}

The velocity at time \(t=30\,\text{s}\) is calculated using the expression

\begin{equation}
\vec{v}_3\,\hat{\textbf{i}}=v_{i,3}\,\hat{\textbf{i}}+a_3(t-t_3)\,\hat{\textbf{i}}.
\end{equation}

Using the numerical values, this becomes

\begin{equation}
\vec{v}_3=10\text{m/s}\,\hat{\textbf{i}}+(2\,\text{m/s}^2)(30\,\text{s}-20\,\text{s})\,\hat{\textbf{i}},
\end{equation}

that is,

\begin{equation}
\label{v3i}
\vec{v}_3=30\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation}

As before, the results given by equations \eqref{x3i} and \eqref{v3i} will be useful when writing the position and acceleration for the fourth region. They’ll serve as the initial position and velocity respectively.

Fourth region: \(30\leq t\leq 40\)

We can finally write the equation of motion for the fourth region, noting that it is a motion with constant velocity.  We also note that a time of \(t_4=30\,\text{s}\) has already passed, and so we write \((t-t_4)\) instead of \(t\) for all equations. Explicitly, we have

\begin{equation}
\vec{x}_4=x_{i,4}\,\hat{\textbf{i}}+v_{i,4}(t-t_4)\,\hat{\textbf{i}},
\end{equation}

where \(\vec{x}_{i,4}=650\,\text{m}\) is the position at time \(t=30\,\text{s}\) (found in the previous segment), \(v_{i,4}=30\,\text{m/s}\) is the velocity at time \(t=30\,\text{s}\) (also found in the previous segment), and \(a_4=0\,\text{m/s}^2\).

The position at time \(t=40\) is then, using the numerical values and equation \eqref{x3i}, the following:

\begin{equation}
\vec{x}_4=650\,\text{m}\,\hat{\textbf{i}}+(30\,\text{m/s})(40\,\text{s}-30\,\text{s})\,\hat{\textbf{i}}.
\end{equation}

The result is

\begin{equation}
\label{x4i}
\vec{x}_4=950\,\text{m}\,\hat{\textbf{i}}.
\end{equation}

The velocity at time \(t=40\,\text{s}\) is known because the car is moving with constant velocity here, and so we can just use the velocity found for the end of the third segment:

\begin{equation}
\label{v4i}
\vec{v}_4=30\,\text{m/s}\,\hat{\textbf{i}}.
\end{equation}

Thus, the total displacement between 0 and 40 seconds is, using equation \eqref{Displace}, and the fact that the initial position is zero, the following:

\begin{equation}
\label{Displace2}
\vec{D}=950\,\text{m}\,\hat{\textbf{i}}-0\,\text{m}\,\hat{\textbf{i}}.
\end{equation}

c) The acceleration between 40 and 50 seconds increases linearly with time. It is not a constant, clearly. Because the acceleration is not constant, we cannot use the usual relations between the kinematic variables as we have done in the previous parts of the problem. To obtain the equations of motion we would have to integrate twice using the definition for the acceleration in terms of position, that is, we would have to integrate twice the following expression:

\begin{equation}
\vec{a}=\frac{d^2x}{dt^2}\,\hat{\textbf{i}}.
\end{equation}

d) For this part, they ask us to sketch a plot of velocity vs time. Using the results of point b), we already know an expression for the velocity \(v\) for each region in terms of the time and other known variables. We can the make graph in figure 1.

e) For the final part, they ask us to sketch a plot of position vs time. We can then use the results of all the position equations in terms of time that we found in b). If we do so, we get the graph in figure 2.

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