Question

An airplane accelerates down a runway at 5.20 m/s2 for 41.7 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

Answer 1

Given:-

• Initial velocity ,u = 5.20m/s²

• velocity ,f = 0m/s

• Time taken ,t = 41.7s

To Find:-

• Distance covered ,s

Solution:-

Firstly we calculate the initial velocity

• v = u+at

Substitute the value we get

→ 0 = u + 5.20×41.7

→ u = 216. 8m/s

Now, Using 3rd Equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = 216.84² + 2×(-5.20) ×s

→ s = 47019.58 - 10.4

→ s = 47009

Therefore the distance travelled by the airplane is 47.09 km

Answer 2

Answer:

An airplane accelerates down a runway at 5.20 m/s2 for 41.7 s until it finally lifts off the ground. The distance traveled before takeoff will be $47.09\ km$

Explanation:

• According  to the first equation of motion,

         $v = u+at$           ...(1)

• According  to the third equation of motion,

         $v^{2}=u^{2}+2 a s$     ...(2)

• Where S is the distance covered by an object, u is the initial velocity, a is the acceleration due to gravity and t is the time taken.

In the question, it is given that,

$a = -5.20\ m/s^2$ (since it accelerates downward)

$t = 41.7\ s$

$v = 0$

Substitute these values into equation(1),

$u = 0-(-5.20\times41.7) = 216.84\ m/s$

Equation (2) becomes,

${}0^{2}=216.84^{2}+2 \times(-5.20) \times s$

$s=47019.58-10.4$

   $=47009\ m$

   $= 47.09 \mathrm{~km}$

• The distance traveled by airplane before takeoff   $= 47.09\ km$