Question

An aeroplane starts from rest with an acceleration of 4m/s² and takes a run for 35 s before taking off. What is the minimum length of runway and with what velocity the plane took off?

Answer 1

Answer:

S = ut+1/2at^2 

     

         s = 0 + 1/2 * 3 * 35 * 35

         s = 1837.5m

v = u + at

   = 0 + 3 * 35

  = 105 m/s.

Explanation:

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Answer 2

Given :-

• Acceleration, a = 4 m/s²

• Time, t = 35s

• Initial velocity, u = 0 m/s

To Find :-

• Minimum length of runway ( S ).

• Velocity the pane took off.

Solution :-

Using the second law of motion we can find the minimum length of runway.

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

$\longrightarrow \sf S = 0 \times 35 + \dfrac{1}{2} \times 4 \times 35 \times 35 \\\\\longrightarrow S= 0+\dfrac{4900}{2} \\\\\longrightarrow S = 2450$

Minimum length of runway = 2450 m

We can find the velocity the plane took off using first and third law of motion.

First law of motion :-

 $\boxed {\bf v = u+at}$

$\longrightarrow \sf v = 0+4 \times 35 \\\\\longrightarrow v = 0+140 \\\\\longrightarrow \bf v = 140$

Third law of motion :-

  $\boxed{\bf v^2= u^2+2as}$

$\longrightarrow \sf v^2 = 0^2+2 \times 4 \times 2450 \\\\\longrightarrow v^2 = 19600\\\\\longrightarrow v = \sqrt{19600} \\\\\longrightarrow \bf v = 140$

Velocity the plane took off = 140 m/s