Question

A jet plane starts from rest with an acceleration of 3ms-2 and makes a run for 35s before taking off. What is the minimum length of the runway and what is the velocity of the jet at take off?.​

Answer 1

Answer:

v=105m/s

s=1837.5 m

Explanation:

a=3m/s²

t=35s

u=0m/s

v=u+at

v=0+3X35

v=105m/s

s=ut+1/2at²

s=0+1/2X3X(35)²

s=1/2X3X1225

s=1/2X3675

s=1837.5m

Answer 2

$\sf{u = 0 m/s} \\\sf{a = 3 m/s^2} \\\sf{t = 35 sec} \\\sf{s = ?}\\\sf{v = ?}$

___

$\sf{s = ut + \dfrac{1}{2} \; at^2}\\\\\sf{= 0 \times \sf{35 }+ \sf{\dfrac{1}{2} \times 3 \times 35^2}\\\\\sf{= 0 \times \sf{35} + \dfrac{1}{2} \times 3 \times 1225}\\\\\sf{ = \sf{\dfrac{\sf{1}}{\sf{2}}} \times \sf{3675}}\\\\\sf{=\dfrac{\sf{3675}}{\sf{2}} = {\underline{\underline{\sf{1837.5 m}}}}$

___

$\sf{v = u + at} \\\sf{= 0 + 3 \times \sf{35}} \\\underline{\underline{\sf{= 105 m/s}}}$

_____