| Composition Setup |
|---|
| Excerpt | ||
|---|---|---|
| ||
Police car with constant acceleration must catch speeder with constant velocity. |
A common type of kinematics problem involves one person or object catching up to another person or object. Here we illustrate a few typical variations of this problem.
| Deck of Cards | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Part B
Again, a speeder driving with a constant speed of 24.5 m/s approaches a (hidden) stationary police car on a long, straight road. The police officer detects the speeder using radar, but this time the police officer begins accelerating at a rate of 5.00 m/s2 only after the speeder has progressed 100 m down the road (the speeder has a 100 m head start). How long will it take for the police car to catch up to the speeder from the instant its acceleration begins? SolutionSystems, Interactions and Models: The same as for Part A.
| |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Wiki Markup | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| A common type of kinematics problem involves one person or object catching up to another person or object. Here we illustrate a few typical variations of this problem.
h4. Part 1
A speeder driving with a constant speed of 24.5 m/s approaches a (hidden) stationary police car on a long, straight road. The police officer detects the speeder using radar, and so the instant the speeder passes the police car, it begins accelerating after the speeder at a rate of 5.00 m/s ^2^. How long will it take for the police car to catch up to the speeder?
Systems: {color:maroon} There are two separate systems here. Both the police car and the speeder's car will be treated as point particles. The police car experiences a constant acceleration as a result of the action of the pavement against its wheels. {color}
Models: {color:maroon}The motion of the speeder's car will be modeled as One-Dimensional Motion with Constant Velocity. The motion of the police car will be modeled as One-Dimensional Motion with Constant Acceleration.{color}
Approach: {color:maroon} We will write an expression for the position of each vehicle as a function of time and then set the positions equal to each other. Since we have the freedom to pick our coordinate system, we can help ourselves in this problem by choosing the positive direction to be the direction of motion and setting _x_ = 0 and t=0 at the point when the speeder passes the police car.
{note}The key to any catch-up problem is that the catch-up occurs when the two systems have the same position.{note}
The appropriate Law of Change for the speeder is:
{latex}\begin{large}\[x_{\rm sp} = x_{\rm sp,i} + v_{\rm sp}t \] \end{large}{latex}
where the subscript "sp" stands for "speeder".
The appropriate Law of Change for the police car is:
{latex}\begin{large}\[ x_{\rm pc} = x_{\rm pc,i} + v_{\rm pc,i} t + \frac{1}{2} a_{\rm pc} t^{2} \]\end{large}{latex}
where "pc" stands for "police car".
Before setting these equations equal to each other, it is a good idea to substitute the zeros so that we can cancel some of the many terms. In this problem, we have chosen our axis in such a way as to make _x_~sp,i~ = _x_~pc,i~ = 0. We also know that _v_~pc,i~ = 0 because the police car was at rest before the speeder passed it.
{note}Although we suggest you solve problems symbolically before substituting numerical values, we make an exception for zeros. Using the zeros of the problem can result in dramatic simplification of the algebra.{note}
Doing this and setting _x_~sp~ = _x_~pc~ gives:
{latex}\begin{large} \[ v_{\rm sp} t = \frac{1}{2} a_{\rm pc} t^{2} \] \end{large}{latex}
This is actually a quadratic equation in _t_, except that the constant term is absent. To be clear, we can rearrange:
{latex}\begin{large} \[ \frac{1}{2} a_{\rm pc} t^{2} - v_{\rm sp} t = 0 \] \end{large}{latex}
This equation has _two_ roots, which are easily obtained by factoring:
{latex}\begin{large} \[ t_1 = 0\] \[ t_{2} = \frac{2 v_{\rm sp}}{a_{\rm pc}} \] \end{large} {latex}
{note}Recognizing that _t_ = 0 is a root is actually a nice confirmation of our model. We have been trying to construct an equation that tells us when the police car and the speeder share the same position. Clearly, they do share the same position at _t_=0 since the speeder was passing the police car. If this root was absent, we should question the validity of our equation!{note}
We select the second root, since we are interested in the time that the police car catches up to the speeder, not the time that the speeder passes the police car. Thus, we find: {color}
{latex}\begin{large} \[ t = \frac{2 v_{\rm sp}}{a_{\rm pc}} = 9.8\;{\rm s} \] \end{large}{latex}
{info}This model could be realistic for speed traps on roads with low speed limits, but it is unlikely to be a good description of a highway chase. Looking at our solution for the time of pursuit, consider how fast the police car will be going when it catches the speeder. Is this realistic for highway speeds?{info}
|
