Equations of Motion

Practice

practice problem 1

Cars cruise down an expressway at 25 m/s. Engineers want to design an interchange for a deceleration of −2.0 m/s2 over 3.0 s.
  1. What velocity will cars have at the end of the approach?
  2. What minimum approach length will satisfy these requirements?
  3. What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)

solution

  1. What velocity will cars have at the end of the approach?
           
      v0 =  25 m/s     v =  v0 + at  
      a =  −2.0 m/s2       v =  (25 m/s) + (−2.0 m/s2)(3.0 s)    
      Δt =  3.0 s     v =  19 m/s  
      v =  ??      
           
  2. What minimum approach length will satisfy these requirements?
           
      v0 =  25 m/s     Δx =  v0t + ½at2  
      a =  −2.0 m/s2     Δx =  (25 m/s)(3.0 s) 
    + ½(−2.0 m/s2)(3.0 s)2  
      Δt =  3.0 s   Δx =  66 m  
      Δx =  ??      
           
  3. What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)
           
      v =  19 m/s     v2 =  v02 + 2aΔx  
      a =  −8.0 m/s2     v02 =  v2 − 2aΔx  
        Δx =  66 m       v02 =  (19 m/s)2 
    − 2(−8.0 m/s2)(66 m)  
      v0 =  ??     v0 =  38 m/s  
           

practice problem 2

A car with an initial velocity of 60 mph needs 144 feet to come to a complete stop. Determine the stopping distance of this same car, under identical conditions with an initial velocity of 30 mph, 20 mph, and 10 mph.

solution

First method …

The hard way to solve this problem is to do it the way that many students think is the easy way. That is, the "plug and chug" method shown in the previous examples. This method appears easy since it requires very little thought, but it turns out to be quite demanding.

First, convert to SI units.

60 mile   1609 m   1 hour  = 26.8  m
1 hour 1 mile 3600 s s
30 mile   1609 m   1 hour  = 13.4  m
1 hour 1 mile 3600 s s
20 mile   1609 m   1 hour  = 8.94  m
1 hour 1 mile 3600 s s
10 mile   1609 m   1 hour  = 4.47  m
1 hour 1 mile 3600 s s
 
144 feet   1 mile   1609 m  = 43.9 m
1 5280 feet 1 mile

Then, calculate acceleration from the data for 60 mph.

  v0 =  26.8 m/s       v2 =  v02 + 2aΔx    
  v =  0 m/s     a =  v2 − v02
  Δx =  43.9 m   2Δx
  a =  ??     a =  −(26.8 m/s)2
    2(43.9 m)
      a =  −8.18 m/s2  

Now, calculate the distances for the other speeds.

  v2 =  v02 + 2aΔx       Δx =  −(13.4 m/s)2  = 11.0 m    
2(−8.18 m/s2)
  Δx =  v2 − v02  =  − v02     Δx =  −(8.94 m/s)2  = 4.89 m  
2a 2a 2(−8.18 m/s2)
    Δx =  −(4.47 m/s)2  = 1.22 m  
2(−8.18 m/s2)

And finally, convert back into English units.

11.0 m   1 mile   5280 feet  = 36 feet
1 1609 m 1 mile
4.89 m   1 mile   5280 feet  = 16 feet
1 1609 m 1 mile
1.22 m   1 mile   5280 feet  = 4 feet
1 1609 m 1 mile

Second method …

Standard problem solving techniques work, but they're a monumental waste of time in this case. Any small error would destroy the answers and waste personal mental energy as well — something we all like to avoid. The easy way to solve this problem does not involve any trickery. It requires that you identify and understand the key concepts needed to solve the problem. Halfway through the mass of equations, an important assumption was made. It was assumed that the braking acceleration of the car would remain constant for all initial velocities. This problem is then one of determining the relationship between displacement and velocity. The equation that does this is

v2 = v02 + 2a(x − x0)

which shows that displacement is proportional to velocity squared when acceleration is constant and either the the initial or final velocity is zero.

x ∝ v2

In this problem we're comparing stopping distances at 30, 20, and 10 mph to those at 60 mph. The square of the ratios of the new velocity to the original velocity will be the ratio of the new stopping distances to the original stopping distance.

The Easier Way to Solve This Problem
ratio of new velocity
to original velocity		 ratio of new distance
to original distance		 new stopping distance
30 mph/60 mph = 1/2 (1/2)2 = 1/4 (1/4) 144 feet = 36 feet
20 mph/60 mph = 1/3 (1/3)2 = 1/9 (1/9) 144 feet = 16 feet
10 mph/60 mph = 1/6 (1/6)2 = 1/36 (1/36) 144 feet = 4 feet

These are exactly the answers we got after doing all the brainless "plug and chug" calculating of the hard method. Isn't it wonderful to have a brain.

practice problem 3

A typical commercial jet airliner needs to reach a speed of 180 knots before it can take off. (A knot is a nautical mile per hour and is very nearly equal to half a meter per second.) If such a plane spends 30 s on the runway estimate …
  1. its acceleration.
  2. the minimum runway length.

solution

  1. To determine acceleration, I recommend using the definition of acceleration.
                 
    v0 =  0 m/s a =  Δv  =  90 m/s  
    v =  180 kts ≈ 90 m/s Δt 30 s
    t =  30 s  a =  3 m/s2 ≈ ⅓ g
    a =  ??          
                 
  2. There are two ways to determine the runway length. Either method yields the same solution.
    • Using the distance-time equation.
             
      v0 =  0 m/s s =  s0 + v0t + ½at2
      v =  180 kts ≈ 90 m/s s =  ½(3 m/s2)(30 s)2
      t =  30 s s =  1350 m
      a =  3 m/s2    
      s =  ??    
             
    • Or from the average velocity during uniform acceleration.
             
      v0 =  0 m/s s =  vt = ½(v + v0)t
      v =  180 kts ≈ 90 m/s s =  ½(90 m/s + 0 m/s)(30 m/s)
      t =  30 s s =  1350 m
      s =  ??    
             

The Physics Hypertextbook

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