Question

starting from rest, an aeroplane takes off after covering 0.7km on the runway. if it takes off at 35m/s, calculate the acceleration and the time for which it moves on the runway.​

Answer 1

GIVEN :

• Distance = 0.7 km = 700 m.
• Final velocity, v = 35 m/s.
• Initial velocity, u = 0.

TO CALCULATE :

• Acceleration, a = ?
• Time, t = ?

FORMULAS USED :

• $\tt v^{2} \ - \ u^{2} \ = \ 2as$
• $\tt a \ = \ \dfrac {v^{2}}{2s}$

SOLUTION :

s = 0.7 km => 700 m.

u = 0

v = 35 m/s.

We know that,

$\implies \sf v^{2} \ - \ u^{2} \ = \ 2as$

$\implies \sf v^{2} \ = \ 2as$

$\implies \sf a \ = \ \dfrac {v^{2}}{2s}$

$\implies \sf a \ = \ \dfrac {35^{2}}{2 \ \times \ 700}$

$\implies \sf a \ = \ 0.875 \ m/s^{2}$

Now, to find time, t = ?

$\implies\sf t \ = \ \dfrac {distance}{speed}$

$\implies \sf t \ = \ \dfrac {700}{35}$

$\implies \sf t \ = \ 20 \ sec$