Putting Equations Together

Summary of Kinematic Equations (Constant a)

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. See an easy reference to the equations needed below.

\(x = x_0 + \bar{v} t\)

\(\bar{v} = \cfrac{v_0 + v}{2}\)

\(v = v_0 + at\)

\(x = x_0 + v_0t + \cfrac{1}{2}at^2\)

\(v^2 = v_0^2 + 2a(x - x_0)\)

Example on Calculating Displacement: How Far Does a Car Go When Coming to a Halt?

On dry concrete, a car can decelerate at a rate of 7.00 m/s², whereas on wet concrete it can decelerate at only 5.00 m/s². Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Strategy

Draw a sketch.

Image credit: OpenStax Physics

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that \(v_0 = 30.0\text{ m/s};\) \(v = 0;\) \(a = -7.00 \mathrm{\; m/s^2}\) (\(a\) is negative because it is in a direction opposite to velocity). We take \(x_0\) to be 0. We are looking for displacement \(\Delta x,\) or \(x - x_0.\)

2. Identify the equation that will help up solve the problem. The best equation to use is

\(v^2 = v_0^2 + 2a(x - x_0).\)

This equation is best because it includes only one unknown, \(x.\) We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for \(x,\) but they require us to know the stopping time, \(t,\) which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for \(x.\)

\(x - x_0 = \cfrac{v^2 \; - \; v_0^2}{2a}\)

4. Enter known values.

\(x \; - \; 0 = \cfrac{0^2 \; - \; (30.0\text{ m/s})^2}{2 \left ( -7.00\mathrm{\; m/s^2} \right )}\)

Thus,

\(x = 64.3\text{ m on dry concrete}.\)

Solution for (b)

This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is \(-5.00\mathrm{\; m/s^2}.\) The result is

\(x_{\text{wet}} = 90.0\text{ m on wet concrete}.\)

Solution for (c)

Once the driver reacts, the stopping distance is the same as it is in parts (a) and (b) for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver’s reaction time.

1. Identify the knowns and what we want to solve for. We know that \(\bar{v} = 30.0\text{ m/s};\) \(t_{\text{reaction}} = 0.500\text{ s};\) \(a_{\text{reaction}} = 0.\) We take \(x_{0 - \text{reaction}}\) to be 0. We are looking for \(x_{\text{reaction}}.\)

2. Identify the best equation to use.

\(x = x_0 + \bar{v} t\) works well because the only unknown value is \(x,\) which is what we want to solve for.

3. Plug in the knowns to solve the equation.

\(x = 0 + (30.0\text{ m/s}) (0.500\text{ s}) = 15.0\text{ m}.\)

This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.

4. Add the displacement during the reaction time to the displacement when braking.

\(x_{\text{braking}} + x_{\text{reaction}} = x_{\text{total}}\)

64.3 m + 15.0 m = 79.3 m when dry
90.0 m + 15.0 m = 105 m when wet

The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time. Image credit: OpenStax, College Physics

Discussion

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.