Question

A plane has a takeoff speed of 94.3 m/s and requires 2345 m to reach that speed. Determine the acceleration of the plane and time required to reach this speed.

Answer 1

a. The acceleration of the plane is equal to 1.90 $m/s^2$.

b. The time required to reach this speed is equal to 49.63 seconds.

Given the following data:

• Initial speed = 94.3 m/s.

• Distance = 2345 meters.

Note: The final speed is equal to zero (0) when the plane reaches its destination (distance).

a. To determine the acceleration of the plane, we would apply the third equation of motion:

What is the third equation of motion?

Mathematically, the third equation of motion is given by this formula:

$V^2 = U^2 - 2aS$

Where:

• V is the final speed.

• U is the initial speed.

• a is the acceleration.

• S is the distance.

Substituting the given parameters into the formula, we have:

$0^2 = 94.3^2 - 2a(2345)\\\\8892.49=4690a\\\\a=\frac{8892.49}{4690}$

Acceleration, a = 1.90 $m/s^2$.

b. To determine the time required to reach this speed:

Mathematically, time is given by this formula:

$t=\frac{V\;+\;U}{a} \\\\t=\frac{0\;+\;94.3}{1.90}$

Time, t = 49.63 seconds.

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