3.4 Motion with constant acceleration  (Page 4/10)

An examination of the equation $v^2=v_0^2+2a(x-x_0)$ can produce additional insights into the general relationships among physical quantities:

• The final velocity depends on how large the acceleration is and the distance over which it acts.
• For a fixed acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance. It takes much farther to stop. (This is why we have reduced speed zones near schools.)

Putting equations together

In the following examples, we continue to explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The note that follows is provided for easy reference to the equations needed. Be aware that these equations are not independent. In many situations we have two unknowns and need two equations from the set to solve for the unknowns. We need as many equations as there are unknowns to solve a given situation.

Before we get into the examples, let’s look at some of the equations more closely to see the behavior of acceleration at extreme values. Rearranging [link] , we have

From this we see that, for a finite time, if the difference between the initial and final velocities is small, the acceleration is small, approaching zero in the limit that the initial and final velocities are equal. On the contrary, in the limit $t\to 0$ for a finite difference between the initial and final velocities, acceleration becomes infinite.

Similarly, rearranging [link] , we can express acceleration in terms of velocities and displacement:

Thus, for a finite difference between the initial and final velocities acceleration becomes infinite in the limit the displacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocities approaches zero for a finite displacement.

How far does a car go?

Strategy

Solution

1. First, we need to identify the knowns and what we want to solve for. We know that v ₀ = 30.0 m/s, v = 0, and a = −7.00 m/s ² ( a is negative because it is in a direction opposite to velocity). We take x ₀ to be zero. We are looking for displacement $\text{Δ}x$ , or x − x ₀ .
   Second, we identify the equation that will help us 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. (Other equations 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.)
   Third, we rearrange the equation to solve for x :
   
   $x-x_0=\frac{v^2-v_0^2}{2a}$
   
   and substitute the known values:
   
   $x-0=\frac{0^2-{(30.0\text{m/s})}^2}{2(\mathrm{-7.00}{\text{m/s}}^2)}.$
   
   Thus,
   
   $x=64.3\text{m on dry concrete}\text{.}$
2. This part can be solved in exactly the same manner as (a). The only difference is that the acceleration is −5.00 m/s ² . The result is
   
   $x_{\text{wet}}=90.0\text{m on wet concrete.}$
3. When the driver reacts, the stopping distance is the same as it is in (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 the velocity remains constant during the driver’s reaction time.
   To do this, we, again, identify the knowns and what we want to solve for. We know that $\stackrel{\text{–}}{v}=30.0\text{m/s}$ , $t_{\text{reaction}}=0.500\text{s}$ , and $a_{\text{reaction}}=0$ . We take $x_{\text{0-reaction}}$ to be zero. We are looking for $x_{\text{reaction}}$ .
   Second, as before, we identify the best equation to use. In this case, $x=x_0+\stackrel{\text{–}}{v}t$ works well because the only unknown value is x , which is what we want to solve for.
   Third, we substitute the knowns to solve the equation:
   
   $x=0+\left(30.0\text{m/s}\right)\left(0.500\text{s}\right)=15.0\text{m}.$
   
   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.
   Last, we then add the displacement during the reaction time to the displacement when braking ( [link] ),
   
   $x_{\text{braking}}+x_{\text{reaction}}=x_{\text{total}},$
   
   and find (a) to be 64.3 m + 15.0 m = 79.3 m when dry and (b) to be 90.0 m + 15.0 m = 105 m when wet.