Question

A jet plane starts from rest with an acceleration of 3 m/s^2 and makes a run for 35 seconds before taking off . What is the displacement covered by the plane ? What is the velocity of the jet as take off ?

Answer 1

Answer:

u=0m/s

a=3m/s^2

t=35 seconds

▪displacement(s)=ut +1/2at^2

=0×35+1/2×3×35^2

=0+1/2×3×1225

=1/2×3675

s= 1837.5m

▪velocity( v)= u+at

= 0+3×35

v=105m/s

HOPE IT HELPS YOU BETTER : )

Answer 2

GIVEN :-

• Initial velocity of jet , u = 0 m/s

• Acceleration , a =3 m/s²

• Time , t = 35 seconds

TO FIND :-

• Displacement covered by the jet

• Velocity of the jet as take off

SOLUTION :-

   Formula to find displacement ,

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

• Displacement , $\sf S = 0\times35+\dfrac{1}{2} \times 3 \times 35 \times 35$

                                      $= \sf 0+\dfrac{3\times1225 }{2}$

                                      $\sf = \dfrac{3675}{2}$

                                      $\bf = 1837.5 \ cm$

    Formula to find velocity ,

     

                     $\bullet\sf \ v=u+at\\\\ \bullet v^2=u^+2as$

   

  Method 1

 

        $\longrightarrow \sf v=0+3\times35\\\\\longrightarrow \bf v = 105 \ m/s$

       

  Method 2

        $\longrightarrow\sf v^2 = 0^2+2\times3\times1837.5 \\\\\longrightarrow v^2 = 11025 \\\\\longrightarrow v =\sqrt{11025 } \\\\\longrightarrow \bf = 105 \ m/s$