Malik walked up and back down an empty 18m long escalator at a constant rate. Since the escalator was going up at the time, it took him only 22.5s. Going down took Malik 90s. Assume the escalator speed is constant. What was Malik's walking speed? (m)/(s)

Define variables: Define the variables for Malik's walking speed and the escalator speed. Let's denote Malik's walking speed as $ v_m $ (in meters per second) and the escalator speed as $ v_e $ (in meters per second).Set up equations: Set up the equations based on the given information. When Malik is going up, the escalator's speed adds to his walking speed, so the total speed is $ v_m + v_e $. It takes him 22.5 seconds to cover the 18 meters, so we have the equation: $ 18 = (v_m + v_e) \times 22.5 $ When going down, the escalator's speed subtracts from his walking speed, so the total speed is $ v_m - v_e $. It takes him 90 seconds to cover the same distance, so we have the equation: $ 18 = (v_m - v_e) \times 90 $Solve first equation: Solve the first equation for $ v_m + v_e $. $ v_m + v_e = \frac{18}{22.5} $ Calculate the right side of the equation. $ v_m + v_e = 0.8 $ m/sSolve second equation: Solve the second equation for $ v_m - v_e $. $ v_m - v_e = \frac{18}{90} $ Calculate the right side of the equation. $ v_m - v_e = 0.2 $ m/sAdd equations: Add the two equations to solve for Malik's walking speed $ v_m $. $ (v_m + v_e) + (v_m - v_e) = 0.8 + 0.2 $ $ 2v_m = 1.0 $ Now, divide both sides by 2 to find $ v_m $. $ v_m = \frac{1.0}{2} $ Calculate the value of $ v_m $. $ v_m = 0.5 $ m/s

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Malik walked up and back down an empty
18m long escalator at a constant rate. Since the escalator was going up at the time, it took him only
22.5s. Going down took Malik
90s. Assume the escalator speed is constant.
What was Malik's walking speed?

(m)/(s)

Malik walked up and back down an empty $18 \mathrm{~m}$ long escalator at a constant rate. Since the escalator was going up at the time, it took him only $22.5 \mathrm{~s}$ . Going down took Malik $90$ s. Assume the escalator speed is constant. $\newline$ What was Malik's walking speed? $\newline$ $\frac{\mathrm{m}}{\mathrm{s}}$

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Q. Malik walked up and back down an empty

18 \mathrm{~m}

long escalator at a constant rate. Since the escalator was going up at the time, it took him only

22.5 \mathrm{~s}

. Going down took Malik

90

s. Assume the escalator speed is constant.

\newline

What was Malik's walking speed?

\newline

\frac{\mathrm{m}}{\mathrm{s}}

Define variables: Define the variables for Malik's walking speed and the escalator speed. $\newline$ Let's denote Malik's walking speed as $v_m$ (in meters per second) and the escalator speed as $v_e$ (in meters per second).
Set up equations: Set up the equations based on the given information. $\newline$ When Malik is going up, the escalator's speed adds to his walking speed, so the total speed is $v_m + v_e$ . It takes him $22$ . $5$ seconds to cover the $18$ meters, so we have the equation: $\newline$ $18 = (v_m + v_e) \times 22.5$ $\newline$ When going down, the escalator's speed subtracts from his walking speed, so the total speed is $v_m - v_e$ . It takes him $90$ seconds to cover the same distance, so we have the equation: $\newline$ $18 = (v_m - v_e) \times 90$
Solve first equation: Solve the first equation for $v_m + v_e$ . $\newline$ $v_m + v_e = \frac{18}{22.5}$ $\newline$ Calculate the right side of the equation. $\newline$ $v_m + v_e = 0.8$ m/s
Solve second equation: Solve the second equation for $v_m - v_e$ . $\newline$ $v_m - v_e = \frac{18}{90}$ $\newline$ Calculate the right side of the equation. $\newline$ $v_m - v_e = 0.2$ m/s
Add equations: Add the two equations to solve for Malik's walking speed $v_m$ . $\newline$ $(v_m + v_e) + (v_m - v_e) = 0.8 + 0.2$ $\newline$ $2v_m = 1.0$ $\newline$ Now, divide both sides by $2$ to find $v_m$ . $\newline$ $v_m = \frac{1.0}{2}$ $\newline$ Calculate the value of $v_m$ . $\newline$ $v_m = 0.5$ m/s