The Physics
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Opus in profectus

Equations of Motion

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Practice

practice problem 1

Cars cruise down an expressway at 25 m/s. Engineers are designing an off-ramp in an interchange with a deceleration of −2.0 m/s2 that lasts 3.0 s.
  1. What velocity will cars have at the end of the off-ramp?
  2. What minimum length should the ramp have?
  3. What maximum velocity could a car entering the off-ramp have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)

solution

  1. State the givens and the unknown. Use the first equation of motion — the one where velocity is a function of time.
       
    v0 =  25 m/s
    a =  −2.0 m/s2
    Δt =  3.0 s
    v =  ?
     
    v = v0 + at
    v = (25 m/s) + (−2.0 m/s2)(3.0 s)
    v = 19 m/s
  2. Restate the givens and the unknown from the previous part, since they're all still valid. Use the second equation of motion — the one where displacement is a function of time.
       
    v0 =  25 m/s
    a =  −2.0 m/s2
    Δt =  3.0 s
    Δs =  ?
     
    Δs = v0t + ½at2
    Δs = (25 m/s)(3.0 s) + ½(−2.0 m/s2)(3.0 s)2
    Δs = 66 m
  3. The final velocity calculated in part a. is still the final velocity. (All cars need to exit at the same speed.) The displacement calculated in part b. is still the displacement. (All cars get the same amount of space to slow down.) The new acceleration is four times the old one. The new initial velocity is the new unknown. No time is given or can be inferred. None is needed. Use the third equation of motion — the one where velocity is a function of displacement and time is not a part of the equation.
       
    v =  19 m/s
    a =  −8.0 m/s2
    Δs =  66 m
    v0 =  ?
     
    v2 = v02 + 2aΔs
    v02 = v2 − 2aΔs
    v02 = (19 m/s)2 − 2(−8.0 m/s2)(66 m)
    2v0 = 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 with an initial velocity of…
  1. 30 mph
  2. 20 mph
  3. 10 mph
Note: The rate of change of velocity is not affected by inital velocity in this problem. Fast cars and slow cars slow down at the same rate.

solution

First method…

The hard way to solve this problem is to do it the way that many students think is the easy way — "numbers in, answers out" or "plug and chug". 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   1,609 m   1 hour  = 26.8 m/s
1 hour 1 mile 3,600 s
30 mile   1,609 m   1 hour  = 13.4 m/s
1 hour 1 mile 3,600 s
20 mile   1,609 m   1 hour  = 8.94 m/s
1 hour 1 mile 3,600 s
10 mile   1,609 m   1 hour  = 4.47 m/s
1 hour 1 mile 3,600 s
 
144 feet   1 mile   1,609 m  = 43.9 m
1 5,280 feet 1 mile

Then calculate the deceleration from 60 mph.

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

Then use this number to calculate the distances for the other speeds.

v2 = v02 + 2aΔs

Eliminate the zero term and solve for displacement.

Δs =  − v02
2a

Numbers in. Answers out.

Δs =  −(13.4 m/s)2  = 11.0 m
2(−8.18 m/s2)
Δs =  −(8.94 m/s)2  = 4.89 m
2(−8.18 m/s2)
Δs =  −(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   5,280 feet  = 36 feet
1 1,609 m 1 mile
4.89 m   1 mile   5,280 feet  = 16 feet
1 1,609 m 1 mile
1.22 m   1 mile   5,280 feet  = 04 feet
1 1,609 m 1 mile

Second method…

Standard problem solving techniques work, but they're a monumental waste of time for this problem. Any small error would destroy the answers and waste personal mental energy, which is something we'd 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 + 2as

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

s ∝ v2

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

v2  ∝  s
                                 

30 mph 2
 = 
1 2
   =    1  =  36 feet        
60 mph 2     4 144 feet        

20 mph 2
 = 
1 2
   =    1  =  16 feet        
60 mph 3     9 144 feet        

10 mph 2
 = 
1 2
   =    1  =  04 feet        
60 mph 6     36 144 feet        

These are the same answers we got using the "plug and chug" method.

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.

    List the givens and unknown first.

    v0 =  0 m/s
    v =  180 kts ≈ 90 m/s
    t =  30 s
    a =  ?

    Then use the equation.

    a =  Δv
    Δt
    a =  90 m/s
    30 s
    a = 3 m/s2 ≈ ⅓ g  
     
  2. There are two ways to determine the runway length. Either method yields the same solution. Let's list the givens and unknown first.
       
    v0 =  0 m/s
    v =  180 kts ≈ 90 m/s
    t =  30 s
    a =  3 m/s2
    s =  ?

    Here's the solution using the distance-time equation.

    s =  s0 + v0t + ½at2
    s =  ½(3 m/s2)(30 s)2
    s = 1350 m

    And here's the solution using the average velocity equation.

    s =  v̅t = ½(v + v0)t
    s =  ½(90 m/s + 0 m/s)(30 m/s)
    s = 1350 m