Question

An aeroplane lands at 216 km h ^-1 and stops after covering a runway of 2km . calculate the acceleration and the time, in which it's comes to rest​

Answer 1

Given :-

• Final velocity (v) = 0 m/s
• Initial velocity (u) = 216 km/h

In m/s :

• 216 km/h = 60 m/s
• Distance (s) = 2 km

In m :

• 2 km = 2000 m

To find :

• Acceleration (a) &
• Time (t)

According to the question,

→ v² = u² + 2as

Where,

• v stands for Final velocity
• u stands for Initial velocity
• t stands for Time
• a stands for Acceleration

→ Substituting the values,

→ (0)² = (60)² + 2 × a × 2000

→ 0 = 3600 + 4000a

→ -3600 = 4000a

→ - 3600 ÷ 4000 = a

→ - 0.9 = a

.°. The acceleration is - 0.9 m/s². The negative signs means that it is retardation.

Now,we will find time by using Newton's first equation of motion

→ v = u + at

Where,

• v stands for Final velocity
• u stands for Initial velocity
• a stands for Acceleration
• t stands for Time

→ Substituting the values,

→ 0 = 60 + (-0.9) × t

→ 0 - 60 = - 0.9t

→ - 60 = - 0.9t

→ 60 ÷ 0.9 = t

→ 66.67 = t

.°. The time is 66.67 seconds..

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