The Physics
Hypertextbook
Opus in profectus

# Equations of Motion

## Practice

### practice problem 1

The speed limit of a particular section of freeway is 25 m/s. The right travel lane is connected to an exit ramp with a short auxiliary lane. Cars would have a comfortable deceleration of −2.0 m/s2 for 3.0 s in the auxiliary lane if they were driving at the speed limit.
1. What speed will cars have when they are done decelerating in this way? (This is also the speed limit of the exit ramp.)
2. What minimum length should the auxiliary lane be to allow for this deceleration?

Drivers don't always drive at the speed limit, and highway engineers take this into consideration.

1. Assume a car could decelerate at four times the "comfortable" rate without losing control. At what maximum speed could a car enter an auxiliary lane with the length you calculated in part b. and still exit at the intended speed?
2. Assume a driver was traveling on the freeway at the speed you calculated in part c. What distance is needed to decelerate this car to the speed limit of the exit ramp at the "comfortable" rate ? #### solution

1. State the givens and the unknown. Use the first equation of motion — the one where speed is a function of time.

 v0 = 25 m/s a = −2.0 m/s2 ∆t = 3.0 s v = ?
 v = v0 + atv = (25 m/s) + (−2.0 m/s2)(3.0 s)v = 19 m/s
2. Restate the givens from the previous part, since they're all still valid, but now is distance is the unknown. Use the second equation of motion — the one where distance is a function of time.

 v0 = 25 m/s a = −2.0 m/s2 ∆t = 3.0 s ∆s = ?
 ∆s = v0t + ½at2∆s = (25 m/s)(3.0 s) + ½(−2.0 m/s2)(3.0 s)2∆s = 66 m
3. The final speed calculated in part a. is still the final speed. (All cars need to exit at the same speed.) The distance calculated in part b. is still the distance. (All cars get the same amount of space to slow down.) The new acceleration is four times the old one. The new initial speed is the new unknown. No time is given or can be inferred. None is needed. Use the third equation of motion — the one where speed is a function of distance and time is not a part of the equation.

 v = 19 m/s a = −8.0 m/s2 ∆s = 66 m v0 = ?
 v2 = v02 + 2a∆sv02 = v2 − 2a∆sv02 = (19 m/s)2 − 2(−8.0 m/s2)(66 m)2v0 = 37.6 m/s
4. The final speed calculated in part a. is still the final speed. (All cars still need to exit at the same speed.) The acceleration is the value given at the start of the problem. (Let's allow the speed demons to accelerate comfortably, too.) The new initial speed is the value we just calculated. The new distance is the new unknown. Again no time is given or can be inferred, so again use the third equation of motion.

 v = 19 m/s a = −2.0 m/s2 v0 = 37.6 m/s ∆s = ?
 ∆v2 = v02 + 2a∆s2∆s = (v2 − v02)/2a2∆s = [(19 m/s)2 − (37.6 m/s)2]/[2(−2.0 m/s2)]2∆s = 264 m

### 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 little thought, but it turns out to be quite demanding.

First, convert to SI units.

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

Then calculate the deceleration from 60 mph.

 v0 = 26.8 m/s v = 0 m/s ∆s = 43.9 m a = ?
v2 =  v02 + 2as

a =
 v2 − v02 2∆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 + 2as

Eliminate the zero term and solve for displacement.

 ∆s = − v02 2a

 ∆s = −(13.4 m/s)2 = 11.0 m 2(−8.18 m/s2) ∆s = −(8.94 m/s)2 = 4.89 m 2(−8.18 m/s2) ∆s = −(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 = 04 feet 1 1609 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 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 = ∆v ∆t
 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 = vt = ½(v + v0)t∆s = ½(90 m/s + 0 m/s)(30 m/s)∆s = 1350 m

### practice problem 4

A 10 car subway train is sitting in a station. It reaches its cruising speed after accelerating at 0.75 m/s2 for distance equivalent to the length of the station (184 m). It then travels at a constant speed towards the next station 18 blocks away (1425 m).
1. Determine the train's cruising speed.
2. Determine the time it took for the train to accelerate from rest to its cruising speed.
3. How long does it take the train to travel the 18 blocks to the next station?
The driver stops the train in the second station in half the distance it took to start it at the first station.
1. What is deceleration of the train in the second station?

#### solution

1. List the givens and the unknown for the train as it departs the station. Pick an appropriate equation of motion. I suggest the velocity-displacement equation, a.k.a. the third equation of motion. Almost no algebra is needed. Put numbers in. Get answer out.

 v0 = 0 m/s a = 0.75 m/s2 ∆s = 184 m v = ?
 v2 = v02 + 2a∆s v2 = 2(0.75 m/s2)(184 m) v = 16.6 m/s

This is about 60 km/h or 37 mph

2. There are several ways to determine the time it took to reach cruising speed.

List only the quantities given in the problem and state the new unknown. Pick a new equation. I suggest the displacement-time equation, a.k.a. the second equation of motion. Some algebra is needed. This is followed by the usual numbers in, answer out.

 v0 = 0 m/s a = 0.75 m/s2 ∆s = 184 m t = ?
 ∆s = v0t + ½at2 t2 = 2∆s/a t2 = 2(184 m)/(0.75 m/s2) t = 22.2 s

We could also use the results of our first calculation and add it to the list of known quantities. Adding the final speed to this list means we could use the velocity-time equation, a.k.a. the first equation of motion. We wouldn't need the displacement of the subway if we did this. We would still need a little bit of algebra, however. (Also, the answer may be slightly different depending on how many digits you saved from your calculation for part a.)

 v0 = 0 m/s v = 16.6 m/s a = 0.75 m/s2 t = ?
 v = v0 + at t = v/a t = (16.6 m/s)/(0.75 m/s2) t = 22.2 s

We could also use every known and calculated quantity plus the two equations for average speed and some algebra. I don't recommend this method, but it works.

 v0 = 0 m/s v = 16.6 m/s a = 0.75 m/s2 ∆s = 184 m t = ?
v =  s  =  v0 + v
t 2
t =
 2∆s v
t =
 2(184 m) 16.6 m/s
t =  22.2 s

3. After the train leaves the station, it travels with a constant speed. That makes for an easy problem. No need to distinguish between initial speed, final speed, and average speed anymore. The speed is just the speed.

 v = 16.6 m/s ∆s = 1425 m ∆t = ?
 v = ∆s/∆t ∆t = v/∆a ∆t = (1425 m)/(16.6 m/s) ∆t = 85.8 s
4. This could be done as a GUESS problem (given, unknown, equation, substitute, solve), but given the way it's worded this is a proportion problem.

The initial and final velocities get switched, which means the sign will change but the absolute value of the change is unchanged. These are our constants. We know nothing about the time. We could make an inference about it, but let's not. Distance is halved and acceleration is the goal. Find an equation with initial and final velocities, acceleration, and distance — but not time.

v2 = v02 + 2as

The two changing quantities are in the same term. Everything else is essentially constant, but with a sign switch. That means that the product of acceleration and distance is also constant. They're inversely proportional.

a ∝ 1s

Half the distance means twice the acceleration. That plus a sign switch gives us…

a = −2(0.75 m/s2)

a = −1.50 m/s2