How to Solve a Lift Journey with Acceleration, Cruising and Braking

An ascending lift starts at rest, speeds up, travels at a steady speed, and then slows down until it stops at the next floor. A student tries to calculate its acceleration using the initial and final velocities of the entire journey. Both are zero, so the calculated acceleration is zero.

The calculation isn’t wrong. The assumption is.

The lift does not have one constant acceleration throughout its journey. It has three distinct stages: acceleration, cruising, and braking. Each stage needs its own calculation.

The reliable method is to split the journey into these three stages, apply the uniform acceleration equations separately, and then combine the distances and times. This is how to solve a lift journey without mixing up acceleration, velocity, and braking.

01Understand the three stages of a lift journey

Imagine travelling in a lift from the ground floor to the fifth floor.

You might notice three things:

1. Just after departure, the lift gradually picks up speed.
2. During the middle of the journey, it travels at approximately constant speed.
3. As it approaches the destination, it slows down until it stops.

These observations describe three different types of motion.

Stage 1: Accelerating

At the beginning, the lift starts from rest and increases its speed.

If the lift is travelling upwards, its acceleration is upwards.

For example, its velocity might increase from \(0\) to \(3.0\,\mathrm{m\,s^{-1}}\) over four seconds.

During this stage:

• Initial velocity is zero if the lift starts from rest.
• Final velocity is greater than zero.
• Acceleration is positive if upwards is positive.
• Velocity increases with time.

Stage 2: Cruising

Once the lift reaches its cruising speed, it travels at a constant velocity.

This is an important distinction: constant velocity does not mean zero velocity.

The lift is still moving upwards, but its acceleration is zero.

During this stage:

• Velocity remains constant.
• Acceleration is zero.
• Displacement is calculated using speed multiplied by time.

Stage 3: Braking

As the lift approaches its destination, it reduces its speed until it reaches zero.

If the lift is still travelling upwards, its acceleration must point downwards.

This means its velocity is positive while its acceleration is negative, assuming upwards is positive.

The lift is still travelling upwards during most of the braking stage. It simply becomes slower.

Notice that negative acceleration does not necessarily mean an object is moving downwards.

It means acceleration is directed downwards. During braking, the lift can still be travelling upwards.

What changes if the lift is descending?

The same principles apply, but the signs reverse.

If upwards is positive:

• A descending lift has negative velocity.
• It accelerates downwards with negative acceleration.
• It cruises with constant negative velocity.
• It brakes with positive acceleration.

The key is to distinguish the direction of velocity from the direction of acceleration.

If you need a refresher on choosing signs in vertical-motion problems, see How to Choose Positive Direction in HSC Vertical Motion.

02Which equations should you use?

Each stage of a lift journey can be treated as a separate uniform-motion problem.

For the accelerating and braking stages, use the three uniform acceleration equations.

The uniform acceleration equations

\[ v=u+at \]

\[ s=ut+\frac{1}{2}at^2 \]

\[ v^2=u^2+2as \]

The symbols represent:

For the cruising stage, acceleration is zero and velocity is constant.

Use:

\[ s=vt \]

Here, \(v\) is the constant cruising velocity.

How do you choose the right equation?

Use this decision table.

You do not need to use every equation in every question.

Choose the equation that contains the quantity you want and the information supplied.

One important rule: reset your variables at each stage

Suppose a lift accelerates from rest until it reaches \(3.0\,\mathrm{m\,s^{-1}}\).

When the lift begins cruising, its initial velocity for that new stage is \(3.0\,\mathrm{m\,s^{-1}}\), not zero.

Likewise, when braking begins, the initial velocity for the braking stage is the cruising velocity.

The final velocity of one stage becomes the initial velocity of the next.

Acceleration can change suddenly in our simplified model, but velocity does not jump instantaneously.

03How can an experiment identify the three stages?

Suppose you wanted to investigate whether a model lift actually has three distinct stages of motion.

A stopwatch would tell you how long the journey takes. But it would not tell you when acceleration finishes or when braking begins.

You need to measure the lift’s position throughout the journey.

Setting up a model-lift investigation

Consider a small lift car travelling vertically along a guide rail.

An ultrasonic motion sensor is fixed at the bottom of the rail, pointing upwards towards a flat reflector attached to the lift car.

The sensor connects to a computer or data logger.

As the lift moves, the sensor sends out sound pulses and measures the time taken for their reflections to return. This allows it to record the distance between the sensor and the lift at regular time intervals.

What should we expect to observe?

If the lift accelerates uniformly, its position should change by increasingly large amounts during equal time intervals.

During cruising, the position should change by equal amounts in equal time intervals.

During braking, the position should continue increasing, but by progressively smaller amounts.

These patterns allow us to identify the three stages.

Illustrative experimental data

The following is an idealised example of the position data that such an investigation could produce.

The lift starts at position zero, accelerates for four seconds, cruises for three seconds, and brakes for four seconds.

These are illustrative model values, not measurements from an actual passenger lift.

Look at the changes in position.

• From 0 to 4 seconds, the lift covers increasingly large distances each second.
• From 4 to 7 seconds, it travels exactly 3 metres each second.
• From 7 to 11 seconds, the distance covered each second decreases.

This is consistent with acceleration, cruising, and braking.

Estimating the acceleration from the observations

During the first four seconds, the lift starts from rest and moves upwards by 6.0 metres.

We can calculate its acceleration using:

\[ s=ut+\frac12at^2 \]

Since \(u=0\):

\[ 6.0=\frac12a(4.0)^2 \]

\[ 6.0=8a \]

\[ \boxed{a=0.75\,\mathrm{m\,s^{-2}}} \]

The lift accelerates upwards at \(0.75\,\mathrm{m\,s^{-2}}\).

Its cruising speed can be estimated from the displacement between 4 and 7 seconds:

\[ v=\frac{15.0-6.0}{3.0} \]

\[ \boxed{v=3.0\,\mathrm{m\,s^{-1}}} \]

Finally, it slows from \(3.0\,\mathrm{m\,s^{-1}}\) to zero over four seconds.

Taking upwards as positive:

\[ a=\frac{v-u}{t} \]

\[ a=\frac{0-3.0}{4.0} \]

\[ \boxed{a=-0.75\,\mathrm{m\,s^{-2}}} \]

The negative sign confirms that the braking acceleration points downwards.

What does this experiment establish?

The position measurements allow us to distinguish the three stages rather than treating the entire journey as one motion.

A stopwatch alone cannot provide that information.

The investigation also has limitations. Real lifts do not usually switch between acceleration, cruising, and braking instantaneously. Their acceleration changes more gradually to make the journey comfortable.

Our uniform acceleration model treats each stage as having a constant acceleration. It is a useful approximation, not a complete description of an actual lift controller.

04A reliable method for solving lift questions

Before attempting a calculation, organise the information into separate stages.

Use this sequence:

Step 1: Choose a positive direction.

For an ascending journey, upwards is usually convenient.

Step 2: Identify the three stages.

Write down when acceleration begins, when cruising starts, and when braking begins.

Step 3: Record the known quantities for each stage.

Keep separate values for initial velocity, final velocity, acceleration, displacement, and time.

Step 4: Calculate the accelerating stage.

Find the velocity at the end of acceleration and the displacement covered.

Step 5: Calculate the cruising stage.

Use the velocity reached at the end of acceleration as the cruising velocity.

Step 6: Calculate the braking stage.

Use the cruising velocity as the initial velocity for braking. Remember that the acceleration points opposite to the direction of travel.

Step 7: Combine the results.

Add the times and signed displacements.

For a lift that travels in one direction without reversing, total distance is the magnitude of its total displacement.

Be careful when solving for time

Sometimes a question gives displacement, initial velocity, and acceleration, and asks you to calculate time.

You may need to rearrange:

\[ s=ut+\frac12at^2 \]

This produces a quadratic equation.

A quadratic can have two roots, but you cannot automatically accept both.

Check:

• Is the time positive?
• Does the time fall within the stage being investigated?
• Is the velocity at that time consistent with the direction of travel?
• Has the lift already stopped before the calculated time?

A root can be mathematically correct but physically irrelevant to the stage you are analysing.

We will examine this properly in the second worked example.

A common mistake: using gravitational acceleration

Do not automatically substitute \(a=-9.8\,\mathrm{m\,s^{-2}}\) just because the lift is moving vertically.

That value describes gravitational acceleration near Earth’s surface during free fall.

A powered lift has its own net acceleration, determined by the forces acting on it. For a kinematics question, use the acceleration given or calculated for the lift.

If a question asks about the forces on the lift, that is a separate dynamics calculation.

05Worked examples

06Questions and solutions

The following questions progress from straightforward three-stage calculations to more demanding problems involving signs, quadratic equations, and competing motion models.

Question 1

A lift descends from rest and travels towards a lower floor.

It accelerates downwards uniformly at \(0.40\,\mathrm{m\,s^{-2}}\) for 5.0 seconds. It then cruises at constant velocity for 6.0 seconds before braking uniformly with an upwards acceleration of \(0.50\,\mathrm{m\,s^{-2}}\) until it stops.

Take upwards as the positive direction.

Calculate:

(a) The velocity at the end of the acceleration stage.

(b) The displacement during each of the three stages.

(c) The total distance travelled and total journey time.

Solution 1

The lift descends a total distance of 21.0 metres in 15.0 seconds. Its velocity after the initial acceleration stage is \(-2.0\,\mathrm{m\,s^{-1}}\).

We take upwards as positive, so downward motion and downward acceleration are negative.

Step 1: Calculate the velocity after acceleration.

The lift starts from rest:

\[ u=0 \]

Its acceleration is:

\[ a=-0.40\,\mathrm{m\,s^{-2}} \]

Time:

\[ t=5.0\,\mathrm{s} \]

Use:

\[ v=u+at \]

\[ v=0+(-0.40)(5.0) \]

\[ \boxed{v=-2.0\,\mathrm{m\,s^{-1}}} \]

The negative sign confirms that the lift is travelling downwards.

Step 2: Calculate the displacement during acceleration.

\[ s=ut+\frac12at^2 \]

\[ s=0+\frac12(-0.40)(5.0)^2 \]

\[ \boxed{s_1=-5.0\,\mathrm{m}} \]

Step 3: Calculate the cruising displacement.

The cruising velocity is \(-2.0\,\mathrm{m\,s^{-1}}\).

The cruising time is 6.0 seconds.

Use:

\[ s=vt \]

\[ s=(-2.0)(6.0) \]

\[ \boxed{s_2=-12.0\,\mathrm{m}} \]

Step 4: Calculate the braking displacement and time.

At the beginning of braking:

\[ u=-2.0\,\mathrm{m\,s^{-1}} \]

At the end:

\[ v=0 \]

The braking acceleration is positive:

\[ a=+0.50\,\mathrm{m\,s^{-2}} \]

Calculate the braking time:

\[ t=\frac{v-u}{a} \]

\[ t=\frac{0-(-2.0)}{0.50} \]

\[ \boxed{t_3=4.0\,\mathrm{s}} \]

Now calculate displacement:

\[ s=ut+\frac12at^2 \]

\[ s=(-2.0)(4.0)+\frac12(0.50)(4.0)^2 \]

\[ s=-8.0+4.0 \]

\[ \boxed{s_3=-4.0\,\mathrm{m}} \]

Notice that the braking acceleration is positive even though the lift is still travelling downwards.

Step 5: Calculate the total distance and time.

The total displacement is:

\[ s_{\text{total}}=-5.0-12.0-4.0 \]

\[ s_{\text{total}}=-21.0\,\mathrm{m} \]

The total distance is the magnitude:

\[ \boxed{d=21.0\,\mathrm{m}} \]

The total time is:

\[ t_{\text{total}}=5.0+6.0+4.0 \]

\[ \boxed{t_{\text{total}}=15.0\,\mathrm{s}} \]

The lift travels 21.0 metres downwards in 15.0 seconds.

Question 2

A lift is descending at \(2.8\,\mathrm{m\,s^{-1}}\) when its braking system engages.

Its acceleration during braking is \(+0.70\,\mathrm{m\,s^{-2}}\), taking upwards as positive.

The destination floor is 5.0 metres below the position where braking begins.

(a) Is the lift speeding up or slowing down?

(b) Calculate the time required to stop.

(c) Calculate the stopping displacement.

(d) Does the lift stop at the destination floor? If not, calculate its velocity as it passes the floor.

Solution 2

The lift is slowing down, but it does not stop at the destination floor. It passes the floor at approximately \(0.92\,\mathrm{m\,s^{-1}}\) downwards and stops 0.60 metres beyond it.

Step 1: Determine whether the lift is speeding up or slowing down.

The initial velocity is:

\[ u=-2.8\,\mathrm{m\,s^{-1}} \]

The acceleration is:

\[ a=+0.70\,\mathrm{m\,s^{-2}} \]

Velocity and acceleration have opposite signs.

Therefore, the lift is slowing down.

This is why positive acceleration does not automatically mean an object is speeding up.

Step 2: Calculate the stopping time.

At rest:

\[ v=0 \]

Using:

\[ v=u+at \]

\[ 0=-2.8+0.70t \]

\[ 0.70t=2.8 \]

\[ \boxed{t=4.0\,\mathrm{s}} \]

Step 3: Calculate the stopping displacement.

Use:

\[ s=ut+\frac12at^2 \]

\[ s=(-2.8)(4.0)+\frac12(0.70)(4.0)^2 \]

\[ s=-11.2+5.6 \]

\[ \boxed{s=-5.6\,\mathrm{m}} \]

The lift needs 5.6 metres to stop.

Step 4: Compare the stopping position with the destination floor.

The destination floor is 5.0 metres below the braking position.

The lift needs 5.6 metres to stop.

Therefore, it travels:

\[ 5.6-5.0=0.60\,\mathrm{m} \]

beyond the destination floor before stopping.

Step 5: Calculate its velocity at the destination floor.

At the floor:

\[ s=-5.0\,\mathrm{m} \]

Use:

\[ v^2=u^2+2as \]

Substitute:

\[ v^2=(-2.8)^2+2(0.70)(-5.0) \]

\[ v^2=7.84-7.00 \]

\[ v^2=0.84 \]

\[ v=\pm\sqrt{0.84} \]

The lift is still descending when it reaches the floor because the stopping displacement has not yet been reached.

Therefore, choose the negative root:

\[ \boxed{v=-0.92\,\mathrm{m\,s^{-1}}} \]

The lift passes the destination floor while still moving downwards at approximately \(0.92\,\mathrm{m\,s^{-1}}\).

This question demonstrates why you must check the physical direction of motion when choosing a square root.

Question 3

A lift travels upwards through a vertical displacement of 24 metres.

It starts from rest, accelerates uniformly for 3.0 seconds, cruises for 6.0 seconds, and then brakes uniformly for 3.0 seconds until it stops.

Calculate:

(a) The maximum velocity reached.

(b) The acceleration during the first stage.

(c) The acceleration during braking.

A student claims that the maximum velocity must be \(2.0\,\mathrm{m\,s^{-1}}\) because the average speed over the journey is \(2.0\,\mathrm{m\,s^{-1}}\).

Explain why this prediction is incorrect.

Solution 3

The maximum velocity is \(2.67\,\mathrm{m\,s^{-1}}\), the acceleration is \(+0.89\,\mathrm{m\,s^{-2}}\), and the braking acceleration is \(-0.89\,\mathrm{m\,s^{-2}}\).

The student’s average-speed calculation is correct, but average speed is not the same as maximum speed.

Step 1: Use the velocity-time graph to calculate displacement.

The area under a velocity-time graph represents displacement.

The total displacement is the sum of three areas:

• The triangle during acceleration.
• The rectangle during cruising.
• The triangle during braking.

Let \(V\) represent the maximum velocity.

The acceleration triangle has a base of 3.0 seconds:

\[ A_1=\frac12(3.0)V \]

The cruising rectangle has a width of 6.0 seconds:

\[ A_2=6.0V \]

The braking triangle also has a base of 3.0 seconds:

\[ A_3=\frac12(3.0)V \]

Step 2: Add the three areas.

The total displacement is 24 metres:

\[ 24=1.5V+6.0V+1.5V \]

\[ 24=9V \]

\[ \boxed{V=2.67\,\mathrm{m\,s^{-1}}} \]

Step 3: Calculate the acceleration.

The lift starts from rest and reaches \(2.67\,\mathrm{m\,s^{-1}}\) in 3.0 seconds.

Use:

\[ a=\frac{v-u}{t} \]

\[ a=\frac{2.67-0}{3.0} \]

\[ \boxed{a=+0.89\,\mathrm{m\,s^{-2}}} \]

Step 4: Calculate the braking acceleration.

During braking, the lift changes velocity from \(+2.67\,\mathrm{m\,s^{-1}}\) to zero over 3.0 seconds.

\[ a=\frac{0-2.67}{3.0} \]

\[ \boxed{a=-0.89\,\mathrm{m\,s^{-2}}} \]

The negative sign indicates that acceleration is directed downwards.

Step 5: Explain the average-speed misconception.

The average speed is:

\[ v_{\text{avg}}=\frac{24}{12} \]

\[ v_{\text{avg}}=2.0\,\mathrm{m\,s^{-1}} \]

However, the lift does not travel at its maximum speed throughout the journey.

It begins at zero, increases to its maximum speed, maintains that speed, and then decreases back to zero.

Consequently, its maximum speed must be greater than its average speed.

The two quantities describe different things.

Question 4

An ascending lift travels a total displacement of 20.0 metres, starting and finishing at rest.

It accelerates uniformly upwards at \(0.80\,\mathrm{m\,s^{-2}}\) for 3.0 seconds.

It then cruises at constant velocity before braking uniformly at \(-0.40\,\mathrm{m\,s^{-2}}\) until it stops.

Calculate:

(a) The cruising velocity.

(b) The displacement during acceleration and braking.

(c) The cruising time.

(d) The total journey time.

Assume the lift has a genuine cruising stage.

Solution 4

The lift cruises for approximately \(3.83\) seconds. Its total journey takes approximately \(12.83\) seconds.

Notice that the acceleration and braking magnitudes are different. This means the two stages do not take the same amount of time.

Step 1: Calculate the cruising velocity.

The lift starts from rest:

\[ u=0 \]

Acceleration:

\[ a=+0.80\,\mathrm{m\,s^{-2}} \]

Time:

\[ t=3.0\,\mathrm{s} \]

Use:

\[ v=u+at \]

\[ v=0+(0.80)(3.0) \]

\[ \boxed{v=2.4\,\mathrm{m\,s^{-1}}} \]

Step 2: Calculate the displacement during acceleration.

\[ s=ut+\frac12at^2 \]

\[ s=0+\frac12(0.80)(3.0)^2 \]

\[ \boxed{s_1=3.6\,\mathrm{m}} \]

Step 3: Calculate the braking time.

The lift begins braking at \(2.4\,\mathrm{m\,s^{-1}}\) and finishes at rest.

\[ u=2.4,\qquad v=0,\qquad a=-0.40 \]

Use:

\[ v=u+at \]

\[ 0=2.4-0.40t \]

\[ \boxed{t_3=6.0\,\mathrm{s}} \]

Braking takes twice as long as the initial acceleration stage because the braking acceleration has half the magnitude.

Step 4: Calculate the braking displacement.

\[ s=ut+\frac12at^2 \]

\[ s=(2.4)(6.0)+\frac12(-0.40)(6.0)^2 \]

\[ s=14.4-7.2 \]

\[ \boxed{s_3=7.2\,\mathrm{m}} \]

Step 5: Calculate the cruising displacement.

The total displacement is 20.0 metres.

Therefore:

\[ s_{\text{cruise}}=20.0-3.6-7.2 \]

\[ \boxed{s_2=9.2\,\mathrm{m}} \]

Step 6: Calculate the cruising time.

The cruising velocity is constant at \(2.4\,\mathrm{m\,s^{-1}}\).

Use:

\[ s=vt \]

Rearrange:

\[ t=\frac{s}{v} \]

\[ t=\frac{9.2}{2.4} \]

\[ \boxed{t_2=3.83\,\mathrm{s}} \]

Step 7: Calculate the total journey time.

\[ t_{\text{total}}=t_1+t_2+t_3 \]

\[ t_{\text{total}}=3.0+3.83+6.0 \]

\[ \boxed{t_{\text{total}}=12.83\,\mathrm{s}} \]

The lift spends approximately 3.0 seconds accelerating, 3.83 seconds cruising, and 6.0 seconds braking.

Question 5

An ascending lift travels 18.0 metres in a total time of 12.0 seconds.

It starts from rest and finishes at rest.

A simplified motion model has three stages:

• Uniform upward acceleration of \(+0.60\,\mathrm{m\,s^{-2}}\).
• A constant-velocity cruising stage.
• Uniform braking until the lift stops.

The acceleration during launch is known, but the magnitude of the braking acceleration has not been measured.

A student claims that the total displacement and journey time are enough to determine the cruising velocity and all three stage durations.

Investigate this claim.

(a) First assume that the braking acceleration is \(-0.60\,\mathrm{m\,s^{-2}}\). Calculate the possible cruising velocities and identify which solution is physically valid.

(b) Calculate the duration of each stage for the physically valid solution.

(c) Now suppose the braking acceleration is instead \(-1.20\,\mathrm{m\,s^{-2}}\). Determine whether the lift could still travel 18.0 metres in 12.0 seconds. If so, calculate the cruising velocity and all three stage durations.

(d) Explain why the original information does not uniquely determine the lift’s motion, and identify one additional measurement that would distinguish the two possible journeys.

Assume that each acceleration is uniform within its stage, the cruising velocity is constant, and there is no waiting time or reversal of direction.

Solution 5

The student’s claim is incorrect. The total displacement and journey time do not uniquely determine the cruising velocity when the braking acceleration is unknown.

In fact, two different physically valid journeys can satisfy the same total displacement and total time.

We will construct both.

Step 1: Define the unknown quantities.

Let:

• \(V\) be the maximum cruising velocity.
• \(a=0.60\,\mathrm{m\,s^{-2}}\) be the launch acceleration magnitude.
• \(b\) be the positive magnitude of the braking acceleration.
• \(t_a\) be the acceleration time.
• \(t_c\) be the cruising time.
• \(t_b\) be the braking time.

The total time is:

\[ t_a+t_c+t_b=12.0 \]

The lift begins and finishes at rest.

During acceleration:

\[ V=at_a \]

Therefore:

\[ t_a=\frac{V}{a} \]

During braking:

\[ V=bt_b \]

Therefore:

\[ t_b=\frac{V}{b} \]

The cruising time must consequently be:

\[ t_c=12-\frac{V}{a}-\frac{V}{b} \]

This relationship is important because the cruising time cannot be negative.

Step 2: Construct a general displacement equation.

During acceleration, the lift starts from rest and reaches velocity \(V\).

Using:

\[ v^2=u^2+2as \]

we obtain:

\[ V^2=2as_a \]

Therefore:

\[ s_a=\frac{V^2}{2a} \]

Similarly, during braking:

\[ s_b=\frac{V^2}{2b} \]

During cruising:

\[ s_c=Vt_c \]

The total displacement is the sum of these three displacements:

\[ 18=\frac{V^2}{2a} +\frac{V^2}{2b}+Vt_c \]

Substituting the expression for cruising time:

\[ 18=\frac{V^2}{2a} +\frac{V^2}{2b} +V\left(12-\frac{V}{a}-\frac{V}{b}\right) \]

Simplifying:

\[ \boxed{ 18=12V-\frac{V^2}{2a}-\frac{V^2}{2b} } \]

This equation connects the total displacement, total time, cruising velocity, launch acceleration, and braking acceleration.

Step 3: Investigate the first model, where braking acceleration is \(-0.60\,\mathrm{m\,s^{-2}}\).

The braking magnitude is:

\[ b=0.60 \]

The launch and braking accelerations have equal magnitudes.

Substitute \(a=b=0.60\):

\[ 18=12V-\frac{V^2}{0.60} \]

Rearrange:

\[ V^2-7.2V+10.8=0 \]

Using the quadratic formula:

\[ V=\frac{7.2\pm\sqrt{7.2^2-4(10.8)}}{2} \]

\[ V=\frac{7.2\pm\sqrt{8.64}}{2} \]

This produces two possible roots:

\[ V_1=5.07\,\mathrm{m\,s^{-1}} \]

\[ V_2=2.13\,\mathrm{m\,s^{-1}} \]

Both roots are positive.

However, this does not mean both are physically possible.

Step 4: Test the larger root.

For:

\[ V=5.07\,\mathrm{m\,s^{-1}} \]

The acceleration time is:

\[ t_a=\frac{5.07}{0.60} \]

\[ t_a=8.45\,\mathrm{s} \]

Because the braking acceleration has the same magnitude, braking also takes:

\[ t_b=8.45\,\mathrm{s} \]

The remaining cruising time would be:

\[ t_c=12.0-8.45-8.45 \]

\[ \boxed{t_c=-4.90\,\mathrm{s}} \]

A negative cruising time is impossible.

The lift would need approximately 16.9 seconds just to accelerate and brake, even before including any cruising.

Therefore, this root must be rejected.

Step 5: Calculate the physically valid first model.

Use the smaller root:

\[ V=2.13\,\mathrm{m\,s^{-1}} \]

The acceleration time is:

\[ t_a=\frac{2.13}{0.60} \]

\[ \boxed{t_a=3.55\,\mathrm{s}} \]

The braking time is identical:

\[ \boxed{t_b=3.55\,\mathrm{s}} \]

The cruising time is:

\[ t_c=12.0-3.55-3.55 \]

\[ \boxed{t_c=4.90\,\mathrm{s}} \]

All three stages have physically possible durations.

The corresponding distances are approximately:

This is a valid lift journey.

Step 6: Investigate the second model, where braking acceleration is \(-1.20\,\mathrm{m\,s^{-2}}\).

Now:

\[ a=0.60,\qquad b=1.20 \]

Substitute these values into the general displacement equation:

\[ 18=12V-\frac{V^2}{2(0.60)} -\frac{V^2}{2(1.20)} \]

Simplify:

\[ 18=12V-\frac{V^2}{1.20} -\frac{V^2}{2.40} \]

\[ 18=12V-1.25V^2 \]

Rearrange:

\[ 1.25V^2-12V+18=0 \]

Using the quadratic formula:

\[ V=\frac{12\pm\sqrt{144-90}}{2.5} \]

\[ V=\frac{12\pm\sqrt{54}}{2.5} \]

The roots are:

\[ V_1=7.74\,\mathrm{m\,s^{-1}} \]

\[ V_2=1.86\,\mathrm{m\,s^{-1}} \]

Again, both roots are positive.

The larger root would produce a negative cruising time, so it must be rejected.

Use the smaller root:

\[ \boxed{V=1.86\,\mathrm{m\,s^{-1}}} \]

Step 7: Calculate the stage durations for the second model.

During acceleration:

\[ t_a=\frac{V}{a} \]

\[ t_a=\frac{1.86}{0.60} \]

\[ \boxed{t_a=3.10\,\mathrm{s}} \]

During braking:

\[ t_b=\frac{V}{b} \]

\[ t_b=\frac{1.86}{1.20} \]

\[ \boxed{t_b=1.55\,\mathrm{s}} \]

During cruising:

\[ t_c=12.0-3.10-1.55 \]

\[ \boxed{t_c=7.35\,\mathrm{s}} \]

All three durations are positive.

The corresponding distances are approximately:

This second journey is also physically valid.

Step 8: Explain what the two solutions tell us.

We have constructed two different journeys that satisfy the original information.

Both journeys cover the same distance in the same time.

However, they have different cruising velocities and different stage durations.

An additional measurement, such as the braking acceleration or the exact time at which cruising begins, would distinguish between them.

For example, the first model begins cruising after 3.55 seconds, while the second begins cruising after 3.10 seconds.

The deeper lesson is that a correct set of equations does not guarantee a unique physical solution. You also need enough information to determine the motion.

06What to remember when solving lift journeys

When you encounter an acceleration, cruising, and braking problem, keep these ideas in mind:

• Split the journey into separate stages before substituting numbers.
• Use the final velocity of one stage as the initial velocity of the next.
• Choose a positive direction and keep every sign consistent.
• Remember that cruising means zero acceleration, not zero velocity.
• During braking, acceleration points opposite to the direction of travel.
• Add the signed displacements to obtain total displacement.
• Add the stage times to obtain total journey time.
• Check every quadratic root against the actual motion and the duration of the relevant stage.
• If a calculated cruising time is negative, the assumed three-stage journey is not physically possible.

The most useful next step is to become comfortable interpreting velocity-time graphs. Once you can recognise the acceleration, cruising, and braking sections, you can often identify the correct equations before doing any calculations. You can also use the area under the graph to check whether your calculated displacement makes sense.

For another application of uniform acceleration, see How to Calculate Acceleration from Speed and Stopping Distance.