Question

1. An air-plane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance travelled before take off. *​

Answer 1

Answer:

Initial speed of airplane u=0 m/s

Acceleration of the plane a=3.2 m/s

2

and time taken t=3.28 s

Using formula, distance covered S=ut+

2

1

at

2

⇒ S=0(32.8)+0.5(3.2)(32.8)

2

=1721.3∼1720m

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Answer 2

Answer :

➥ The distance traveled by the airplane before taking off = 1721.344 m

Given :

➤ Intial velocity of airplane (u) = 0 m/s

➤ Acceleration of airplane (a) = 3.20 m/s²

➤ Time taken (t) = 32.8 sec

To Find :

➤ Distance traveled by the airplane before taking off = ?

Solution :

The distance travelled before take off.

From second equation of motion

$\tt{: \implies s = ut + \dfrac{1}{2}a {t}^{2} }$

$\tt{: \implies s = 0 \times 32.8 + \dfrac{1}{2} \times 3.20 \times 32.8 \times 32.8}$

$\tt{: \implies s = 0 + \dfrac{1}{\cancel{\:2\:}} \times \cancel{3.20} \times 32.8 \times 32.8 }$

$\tt{: \implies s = 0 + 1 \times 1.6 \times 32.8 \times 32.8}$

$\tt{: \implies s = 0 + 1.6 \times 32.8 \times 32.8}$

$\tt{: \implies s = 0 + 52.48 \times 32.8}$

$\tt{: \implies s = 0 + 1721.344}$

$\tt{: \implies \green{\underline{\overline{\boxed{\purple{\bf{ \: \: s = 1721.344 \: m \: \: }}}}}}}$

Hence, the distance traveled by the airplane before taking off is 1721.344 m.

⠀

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as