The Physics
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Opus in profectus

# 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 that lasts 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 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. What minimum approach length will satisfy these requirements?
 v0 = 25 m/s a = −2.0 m/s2 Δt = 3.0 s Δx = ??
 Δx = v0t + ½at2 Δx = (25 m/s)(3.0 s) + ½(−2.0 m/s2)(3.0 s)2 Δx = 66 m
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 a = −8.0 m/s2 Δx = 66 m v0 = ??
 v2 = v02 + 2aΔx v02 = v2 − 2aΔx 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. 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 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 acceleration from the data for 60 mph.

 v0 = 26.8 m/s v = 0 m/s Δx = 43.9 m a = ??
 v2 = v02 + 2aΔx a = v2 − v02 2Δx 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 = v2 − v02 = − v02 2a 2a
 Δx = −(13.4 m/s)2 = 11.0 m 2(−8.18 m/s2) Δx = −(8.94 m/s)2 = 4.89 m 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 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 in this case. 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 + 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 ratio of the new velocity to the original velocity will be the ratio of the new stopping distance to the original stopping distance.

ratio of new velocity
to original velocity
ratio of new distance
to original distance
new stopping distance
30 mph/60 mph = 12 (12)2 = 14 (14) 144 feet = 36 feet
20 mph/60 mph = 13 (13)2 = 19 (19) 144 feet = 16 feet
10 mph/60 mph = 16 (16)2 = 136 (136) 144 feet = 04 feet
The easier way to solve this problem

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 v = 180 kts ≈ 90 m/s t = 30 s a = ??
 a = Δv = 90 m/s Δt 30 s a = 3 m/s2 ≈ ⅓ g
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 v = 180 kts ≈ 90 m/s t = 30 s a = 3 m/s2 s = ??
 s = s0 + v0t + ½at2 s = ½(3 m/s2)(30 s)2 s = 1350 m
• Or from the average velocity during uniform acceleration.  s = v̅t = ½(v + v0)t s = ½(90 m/s + 0 m/s)(30 m/s) s = 1350 m