Answer to Question #94199 in Classical Mechanics for You

Question

If an airplane accelerated down a runway at 3.2 m/s squared for 32.8 seconds until it lifts of the ground what is the distance before takeoff?

Answer 1

The distance traveled is given by

$X_{f}=X_{o}+V_{ox}*t+\frac{1}{2}a_{x}t^{2}$

Where:

• Initial position $X_{o}=0m$
• Final position $X_{f}$
• initial velocity $V_{ox}=0\frac{m}{s}$
• Acceleration $a=3.2\frac{m}{s^{2}}$
• time $t=32.8s$

Numerically evaluating

$X_{f}=0m+0\frac{m}{s}*32.8s+\frac{1}{2}*3.2\frac{m}{s^{2}}*(32.8s)^{2}\\ X_{f}=1721m$

The distance traveled before taking off is

$\boxed{X_{f}=1721m}$