An airplane lands and starts down the runway at a southwest velocity of 44 m/s. What constant acceleration allows it to come to a stop in 1.3 km?
To find the constant acceleration required for the airplane to come to a stop, we can use the equations of motion.
The formula that relates distance, initial velocity, acceleration, and time is:
𝑑 = 𝑣0𝑡 + 0.5𝑎𝑡^2
Where:
𝑑 = distance
𝑣0 = initial velocity
𝑎 = acceleration
𝑡 = time
In this case, the airplane starts down the runway with an initial velocity of 44 m/s and we want to find the acceleration that will allow it to come to a stop in a distance of 1.3 km (or 1300 meters).
Substituting the given values into the equation:
1300 = 44 * t + 0.5 * a * t^2
Simplifying the equation:
0.5 * a * t^2 + 44 * t - 1300 = 0
This is a quadratic equation, where the unknown is acceleration ('a'). We need to solve for 'a'.
To solve the quadratic equation, we can use the quadratic formula:
𝑎 = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐)) / 2𝑎
In this case, the coefficients are:
𝑎 = 0.5
𝑏 = 44
𝑐 = -1300
Plugging in these values into the quadratic formula:
𝑎 = (−44 ± √(44^2 − 4 * 0.5 * -1300)) / 2 * 0.5
𝑎 = (−44 ± √(1936 + 2600)) / 1
𝑎 = (−44 ± √(4536)) / 1
Now we calculate the two possible values for acceleration:
𝑎 = (-44 + √4536) / 1 ≈ 15.93 m/s^2
𝑎 = (-44 - √4536) / 1 ≈ -59.93 m/s^2
Since acceleration cannot be negative in this context (it represents the deceleration of the airplane), we discard the negative value.
Therefore, the constant acceleration required for the airplane to come to a stop is approximately 15.93 m/s^2.