Question

an aircraft lands on the runway with a velocity of 80m/s and decelerates at 20m/s^2 to a velocity of 40m/s. Calculate the distance travelled on the runway

Answer 1

QUESTION:-

an aircraft lands on the runway with a velocity of 80m/s and decelerates at 20m/s^2 to a velocity of 40m/s. Calculate the distance travelled on the runway

EXPLANATION:-

• Initial velocity(u)=80 m/s
• Final velocity(v)=40 m/s
• Acceleration= -20 m/s²
• Distance travelled=?

We can find the distance travelled from landing with 80 m/s to slowing down to 40 m/s, using the relation;

v²=u²+2as

(40)²=(80)²+2(-20)s

1600=6400-40a

1600-6400=-40s

-4800=-40s

s=(-4800)/(40)

s=2129 m

The distance travelled = 120 m

Answer 2

Question

An aircraft lands on the runway with a velocity of 80m/s and decelerates at 20m/s² to a velocity of 40m/s. Calculate the distance travelled on the runway.

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Answer

Initial Velocity (u) = 80 m/s

Final Velocity (v) = 40 m/s

Acceleration (a) = -20 m/s² (Since the aircraft decelerates, the acceleration will be negative)

Distance travelled (s) = ?

We'll apply the third equation of motion for the given question.

3rd Equation of Motion ⇒ v² - u² = 2as

Substitute the variables and find the value of 's' (Distance)

⇒ (40)² - (80)² = 2(-20)(s)

⇒ 1600 - 6400 = -40s

⇒ -4800 = -40s

⇒ s = -4800 ÷ -40

⇒ s = 120 m

∴ The total distance travelled on the runway is 120 m.

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Additional Information

Acceleration is the rate of change of velocity in a particular given time.

We have 3 Equations of Motion. They are -:

1st Equation of Motion → v = u + at

2nd Equation of Motion→ $s = ut+\dfrac{1}{2}at^{2}$

3rd Equation of Motion → v² - u² = 2as

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