PHYSICS

Classical Mechanics Level 2

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff


The answer is 1721.344.

Rahul Parmar by RAHUL PARMAR , 22, India

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4 solutions

Nafees Zakir
Oct 3, 2014

s = u t + a t 2 2 = 0 + 3.2 × ( 32.8 ) 2 2 [ W h e n u = 0 m / s , a = 3.2 m / s 2 , t = 32.8 ] = 1721.344 s = ut + \dfrac{at^2}{2} \\ = 0 + \dfrac{3.2\times(32.8)^2}{2} \quad [When \enspace u = 0m/s, a = 3.2m/s^2, t = 32.8] \\ = \boxed{1721.344}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I didn't understand that. a=∆V/∆d,then whatt??

Les Play Gamer - 5 years, 1 month ago
Ana Colon
Nov 23, 2014

d= distance, a= acceleration, t= time

d= 1/2 a t ^ 2

= 1/2 ( 3.2 ) ( 32.8 ^ 2 )

= ( 1.6 ) ( 1075.84 )

= 1721.344 meters

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Krishna Karthik
Nov 21, 2018

d = v i t + a t 2 2 \Large d=v_it+\frac{at^2}{2}

Plug in to get d = 1721.344 d=1721.344

0 Helpful 0 Interesting 0 Brilliant 0 Confused

s=1/2at^2 s=1/2(3.2)(32.8)^2 s=1721.344

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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