Question

an aeroplane starts from rest with an acceleration of 3ms -2 square and takes a run for 35s before taking off. what is the minimum length of the runway and with what velocity does the plane take off

Answer 1

Answer:

v=u+at

v=0+3×35

v=105m/s is the answer

Answer 2

$\large{ \sf \underline{Solution- }}$

Here it is given that,

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 3 m/s²
• Time (t) = 35 sec

➢ We have to find the length of runway [distance] and final velocity of Plane.

Now,

$\implies \: \boxed{ \sf{ \ s= ut+\dfrac{1}{2}at^2}}$

Here,

• s = distance
• u = initial velocity
• a = acceleration
• t = time

$\implies \: \sf{ \ s= ut+\dfrac{1}{2}at^2}$

$\sf\implies \: s= 0\times (35) +\dfrac{1}{2}\times (3)\times (35)^2$

$\sf \implies \: s= 0+ \dfrac{3\times 1225}{2}$

$\sf \implies \: s= \cancel{\dfrac{3675}{2}}$

$\sf \implies \: s = 1837.5 \: m$

➢Length of runway [distance] = 1837.5 m

Now,

$\implies \: \sf \boxed{ \sf v = u + at}$

$\implies \: \sf v = 0 + 3 \times 35$

$\implies \: \sf v = 105\:m/s$

➢Final velocity of Plane (v) = 105 m/s