Application of TSD in Elevator: Questions, Examples

Time, Speed, and Distance (TSD) concepts are not only applicable to trains, boats, and streams but also to escalators or elevator problems. Understanding relative speed is crucial for solving these problems on elevators or escalators, especially in scenarios where moving platforms affect travel times and distances. An escalator is a moving staircase transporting people between different levels. Relative speed in this case will be the speed observed from the perspective of a moving platform or staircase.

The concept of Escalator

The basic things we need to remember when dealing with problems related to escalators are as follows:

• The distance is in terms of the number of steps.

• The total number of steps will always be constant in every situation.

• If we move along with the escalator, we would have to climb fewer steps on our own because the escalator also will push us forward some number of steps on its own. For example, if we climb 50 stairs on our own, and the escalator pushes out 20 stairs in our favour at the same time, then we would have climbed (50 + 20) = 70 stairs in total.

The total number of steps = steps climbed by oneself + steps produced by the escalator.

• If we are moving against the escalator, we would have to climb more steps on our own as the escalator will try to pull us back some number of steps on its own. For example, if we climb 50 stairs on our own, and the escalator pushes out 30 stairs against us at the same time, then we would have climbed (50 - 30) = 20 stairs in total.

The total number of steps = steps climbed by oneself - steps produced by the escalator.

Types of Problems with Escalator

Questions asked from this topic can be of the following types:

Problems based on Time

Example: A man is going up the escalator. It takes him 80 seconds to walk up the escalator which is moving upwards and 120 seconds to walk down the escalator which is moving downwards. Calculate the time taken by Bunty to climb the stationary escalator.

Solution: Let the man’s speed be ‘m’ steps per second.

Let the escalator's speed be ‘e’ steps per second.

Number of steps (N) while walking upward = 80m + 80e (Since the man is moving upwards, 80e will be added)

Again number of steps while walking downward = 120m - 120e (Since the man is moving upwards, 120e will be subtracted)

By equating the number of steps,

80m + 80e =120m - 120e

⇒ 40m = 200e

⇒ m = 5e

So, N = 80m + 80e = 80m + 16m = 96m

Hence, the required time, when the escalator is stationary = $\frac{96m}{m}$ = 96 seconds.

Problems based on Length

Example: When Rik walks down, he takes 50 seconds on an escalator that is moving down, but he takes 40 seconds when he runs down. He takes 10 steps when he is walking, whereas he takes 30 steps when he is running. Find the length of the escalator number in terms of steps.

Solution: Let the speed of the escalator be “e” steps per second.

Now, the distance covered when Rik is walking = (10 + 50e) steps

Also, the distance covered when Rik is running = (30 + 40e) steps

According to the question,

10 + 50e = 30 + 40e

⇒ 10e = 20

⇒ e = 2 steps per second

So, the total number of steps = 10 + 50(2) = 110.

Hence, the length of the escalator is 110 steps.

Problems based on Speed

Example: A man walks up an ascending escalator at a speed of 3 steps per second and reaches the top in 30 steps. If he walks up the escalator at a speed of 5 steps per second, he reaches the top in 45 steps. Find the speed of the escalator.

Solution: Let the speed of the escalator be ‘e’ steps/sec.

At a speed of 3 steps/sec to reach 30 steps, the man took $\frac{30}{3}$ = 10 sec

Distance covered by the escalator in 10 sec = 10e steps.

So, the total distance travelled = (30 + 10e) steps

Again, at a speed of 5 steps/sec to reach 45 steps, the man took $\frac{45}{5}$ = 9 sec

Distance covered by the escalator in 9 sec = 9e steps.

So, the total distance travelled = (45 + 9e) steps

Now, 30 + 10e = 45 + 9e

⇒ 10e - 9e = 45 - 30

⇒ e = 15 steps/sec

Hence, the speed of the escalator is 15 steps/sec.

Problems based on the number of steps

Example: Two persons A and B start descending on an escalator which is going down. B is thrice as fast as A. By the time they reach the bottom, A descends 20 steps while B descends 30 steps. Find the number of visible steps of the escalator.

Solution: Let the escalator take ‘$m$’ steps when A descends 20 steps.

Since B is thrice as fast as A, A would have taken 10 steps by the time B takes 30 steps.

In that time the escalator would have taken $\frac{m}{2}$ steps.

According to the question,

20 + $m$ = 30 + $\frac{m}{2}$

⇒ $m$ - $\frac{m}{2}$ = 30 - 20

⇒ $\frac{m}{2}$ = 10

⇒ $m$ = 20

Hence, the number of visible steps = 20 + 20 = 40.

List of formulas

Tips and Tricks

• Understand the basic concept of relative speed.

• When moving with the escalator,

The total number of steps = steps climbed by oneself + steps produced by the escalator.

• When moving against the escalator,

The total number of steps = steps climbed by oneself − steps produced by the escalator.

• The total number of steps will always be constant in every situation.

Practice Questions/Solved Examples

Q.1.

A man and his wife walk up a moving escalator. The man walks twice as fast as his wife. When he arrives at the top, he has taken 28 steps. When she arrives at the top, she has taken 21 steps. How many steps are visible in the escalator at any one time?

1. 40

2. 42

3. 32

4. 45

Solution:

Since the man’s wife is slower, she would have taken more steps if they were moving against the escalator.

Therefore, they are moving with the escalator, which is going upwards.

Let the wife’s speed be $x$ and the man’s speed be $2x$.

Let the speed of the escalator be $y$.

Since the total number of steps is equal in both cases.

$28+\frac{28}{2x}\times y=21+\frac{21}{x}\times y$

⇒ $x=y$

So, the total visible steps = 28 + $\frac{28}{2y}\times y$ = 28 + 14 = 42

Hence, the correct answer is option (2).

Q.2.

Ripan and Kushal are climbing on a moving escalator that is going up. Ripan takes 60 steps to reach the top but Kushal takes 64 steps to reach the top. Ripan can take 3 steps in a second while Kushal can take 4 steps in a second. Calculate the total number of steps in the escalator.

1. 80

2. 70

3. 85

4. 95

Solution:

Let the speed of the escalator be ‘$e$’ steps/sec.

Now, Ripan takes 60 steps to reach the top and he can take 3 steps in a second.

So, in that time Ripan takes to reach the top, the escalator takes $\frac{60}{3}\times e$ = $20e$ steps.

Thus total steps = $60+20e$

Similarly for Kushal, total steps = $64+\frac{64}{4}\times e$ = $64+16e$

Since the total steps are constant,

So, $60+20e=64+16e$

⇒ $4e=4$

⇒ $e=1$

Therefore, the total number of steps in the escalator = 60 + 20(1) = 80.

Hence, the correct answer is option (1).

Q.3.

A woman is walking down a downward-moving escalator and steps down 10 steps to reach the bottom. Just as she reaches the bottom of the escalator, a sale commences on the floor above. She runs back up the downward-moving escalator at a speed five times that which she walked down. She covers 25 steps in reaching the top. How many steps are visible on the escalator when it is switched off?

1. 15

2. 22

3. 20

4. 25

Solution:

Let the woman’s speed be $w$ with the escalator and $5w$ against the escalator.

Let the speed of the escalator be $e$.

Then, the total number of steps is equal in both cases.

$10+\frac{10}{w}\times e=25-\frac{25}{5w}\times e$

⇒ $w=e$

So, the total length = 10 + $\frac{10}{e}\times e$ = 10 + 10 = 20

Hence, the correct answer is option (3).

Q.4.

A famous mathematician Fibonacci, is always in a hurry and walks up an up-going escalator at the rate of one step per second. Twenty steps bring him to the top. The next day he goes up at two steps per second, reaching the top in 32 steps. How many steps are there in the escalator?

1. 60

2. 70

3. 85

4. 80

Solution:

Let the speed of the escalator be $e$ steps/sec.

Total steps when going at 1 step/sec = $20+\frac{20}{1}\times e$ = $20+20e$

Total steps when going at 2 steps/sec = $32+\frac{32}{2}\times e$ = $32+16e$

According to the question,

$20+20e=32+16e$

⇒ $4e=12$

⇒ $e=3$ steps/sec

So, the total number of steps in the escalator = $20+20\times 3=80$

Hence, the correct answer is option (4).

Q.5.

A man can walk up a moving ‘up’ escalator in 30 sec. The same man can walk down this moving-up escalator in 90 sec. Assume that his walking speed is the same upwards and downwards. How much time will he take to walk up the escalator, when it’s not moving?

1. 45

2. 40

3. 42

4. 44

Solution:

Let the speed of the man be $x$ and the speed of the escalator be $y$.

When the escalator is moving up, the effective speed = $x+y$

When the escalator is not moving, the effective speed = $x$

When the escalator is moving up, but the man is moving downwards the effective speed

= $x-y$

Now, $x+y,x,x-y$ these speeds are in AP, so the times taken in each case are in HP.

Therefore the time taken when the escalator is not moving is the harmonic mean of 30 and 90.

The HM of 30 sec and 90 sec = $\frac{2\times 30\times 90}{30+90}$ = 45 sec

Hence, the correct answer is option (1).

Q.6.

Ripan always walks down a moving escalator to save time. He takes 50 steps while he goes down. One day due to a power failure of 10 seconds (when the escalator comes to a halt) he takes 9 secs more than usual time to get down. Find the visible steps of the escalator.

1. 400

2. 450

3. 500

4. 550

Solution:

Let Ripan cover $n$ stairs in those 10 seconds, after which rest of the remaining distance he climbs down in a normal manner once the escalator starts.

So, climbing down those $n$ stairs with escalator ON and without escalator ON brings a difference of 9 seconds.

Therefore, he normally covered the distance in (10 – 9) = 1 second every day.

Let the speed of Ripan be $v$ and the speed of the escalator be $e$.

So, $\frac{v+e}{v}=\frac{10}{1}$ or $\frac{e}{v}=9$

Therefore, the speed of the escalator is 9 times the speed of Ripan.

Every day he climbs down 50 stairs.

Therefore, in that time the escalator pushes out $(50\times 9)=450$ stairs.

Therefore, the total number of visible steps = 50 + 450 = 500.

Hence, the correct answer is option (3).

Q.7.

Mina is climbing up the moving escalator that is going up. She takes 60 steps to reach the top while Tina is coming down the same escalator. The speed ratio of Mina and Tina is $3:5$. Calculate the number of steps, if it’s given that both of them take the same time to reach the other end.

1. 60

2. 70

3. 80

4. 90

Solution:

Let the speed of Mina and Tina be $3s$ steps/sec and $4s$ steps/sec, respectively.

Let $t$ be the time both take to reach the other end and the speed of the escalator be $e$ steps/sec.

Now, the total number of steps from Mina’s point of view = $3st+et$

Also, the total number of steps from Tina’s point of view = $5st-et$

Since the total number of steps is constant,

So, $5st-et=3st+et$

⇒ $2st=2et$

⇒ $s=e$

When Mina takes 60 steps to reach the top,

The escalator would have taken $\frac{60e}{3s}$ = $\frac{60e}{3e}$ = 20 steps

So, the total number of steps = (60 + 20) = 80.

Hence, the correct answer is option (3).