Question

An aeroplane accelerated down a runway at 3.2m/s for 32.8 seconds until it finally lifts off the ground. The distance travelled before the takeoff?



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Answer 1

Answer:

1721.344 m

Explanation:

By the equations of motion

$x(t) = x(0) + v(0)t + 0.5a {t}^{2}$

so x(t)=0+0+0.5(3.2)(32.8×32.8)

= 1721.344m

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Acceleration of the plane (a) = 3.2 m/s².
• Time taken (t) = 32.8 seconds.
• Initial velocity (u) = 0 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\large{\tt \leadsto S = 0 \times t + \dfrac{1}{2} \times 3.2 \times (32.8)^2}$

$\large{\tt \leadsto S = 0 + \dfrac{1}{2} \times 3.2 \times (32.8)^2}$

$\large{\tt \leadsto S = \dfrac{1}{2} \times 3.2 \times (32.8)^2}$

$\large{\tt \leadsto S = \dfrac{1}{\cancel{2}} \times \cancel{3.2} \times (32.8)^2}$

$\large{\tt \leadsto S = 1 \times 1.6\times (32.8)^2}$

$\large{\tt \leadsto S = 1 \times 1.6\times 1075.84}$

$\large{\tt \leadsto S = 1 \times 1721.344}$

$\huge{\boxed{\boxed{\tt S = 1721.3 \: m}}}$

So, the Distance travelled before Take off is 1721.3 meters.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Additional formulas:-

• $\large{\tt v = u + at}$
• $\large{\tt S = ut + \frac{1}{2}at^2}$
• $\large{\tt v^2 - u^2 = 2as}$
• $\large{\tt S_n = u + \frac{a}{2}(2n - 1)}$

$\rule{300}{1.5}$