An airplane from rest accelerates on a runway at 5.50 m/s2 for 20.25s until it finally takes off the ground.
What is the distance covered before take off?
d=(0m/s)(20.25s) + 1/2 (5.50m/s squared)(20.25s)squared
d= 1128m
d=1130m
a car starts from rest and accelerates uniformly over a time of 5.21 s for a distance of 140 m. Determine the acceleration of the car.
Where's the solution
d = 0.5a*t^2.
Well, if the airplane is accelerating on the runway, it's like a race between the airplane and the ground. Let's call it the "Runway Grand Prix"!
In this Grand Prix, the airplane starts from rest and accelerates at a speedy 5.50 m/s² for a total of 20.25 seconds. So, to find the distance covered before takeoff, we'll need to use the equation:
distance = (initial velocity × time) + (0.5 × acceleration × time²)
Since the airplane starts from rest (initial velocity = 0), we can simplify the equation to:
distance = 0.5 × acceleration × time²
Plugging in our values, we get:
distance = 0.5 × 5.50 m/s² × (20.25 s)²
After doing some number crunching, we find that the distance covered before takeoff is approximately 1115.34 meters. That's quite a run for an airplane!
To find the distance covered by the airplane before takeoff, we can use the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 0 m/s (since the airplane starts from rest)
Acceleration (a) = 5.50 m/s^2
Time (t) = 20.25 s
Plugging in these values into the equation, we can calculate the distance:
distance = 0 * 20.25 + (1/2) * 5.50 * (20.25)^2
Calculating this expression step by step:
distance = 0 + (1/2) * 5.50 * 410.0625
distance = 0 + 1.375 * 410.0625
distance = 562.66884375
Therefore, the distance covered by the airplane before takeoff is approximately 562.67 meters.