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An airplane starts from rest and accelerates at a constant 4.50 m/s² for 35.0 s before leaving the fast was it going when it took off?

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Final answer:

The airplane was moving at a speed of 157.5 m/s when it took off, calculated using the formula for final velocity by plugging in the given constant acceleration of 4.50 m/s² and the time span of 35.0 seconds.

Step-by-step explanation:

Calculating the Takeoff Speed of an Airplane

An airplane starts from rest and accelerates down a runway with a constant acceleration. To find the speed when the airplane took off, we use the formula for calculating final velocity when initial velocity, acceleration, and time are known:

Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))

Since the airplane starts from rest, its initial velocity (u) is 0 m/s. Given the constant acceleration of 4.50 m/s² and the time span of 35.0 seconds before takeoff, we can plug these values into the formula to calculate its takeoff speed:

Final velocity = 0 m/s + (4.50 m/s² × 35.0 s)

By multiplying the acceleration by the time, we get:

Final velocity = 4.50 m/s² × 35.0 s = 157.5 m/s

This is the speed at which the airplane was moving when it took off.

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