A certain aircraft has a mass of 300,000 kg.At a certain instant during its landing,its speed is 27.0 m/s.If the braking force is a constant 445,000 N,what is the speed of the airplane 10.0 s later?
A)10.0 m/s
B)12.2 m/s
C)14.0 m/s
D)18.0 m/s
E)20.0 m/s

Question

Quizplus · Accepted Answer

To find the speed of the airplane 10.0 s later, we use the formula derived from Newton's second law, $F = ma$, where $F$ is the force applied, $m$ is the mass of the object, and $a$ is the acceleration. First, we find the acceleration using the formula $a = \frac{F}{m}$. Given that $F = -445,000$ N (negative because it's a braking force, opposing the motion) and $m = 300,000$ kg, we have:$$a = \frac{-445,000 \, \text{N}}{300,000 \, \text{kg}} = -1.483 \, \text{m/s}^2$$Next, we use the formula $v = u + at$ to find the final velocity ($v$), where $u$ is the initial velocity (27.0 m/s), $a$ is the acceleration we just calculated, and $t$ is the time (10.0 s). Plugging in the values:$$v = 27.0 \, \text{m/s} + (-1.483 \, \text{m/s}^2)(10.0 \, \text{s}) = 27.0 \, \text{m/s} - 14.83 \, \text{m/s} = 12.17 \, \text{m/s}$$Rounding to the nearest tenth gives us 12.2 m/s, which corresponds to option B.

A Certain Aircraft Has a Mass of 300,000 Kg