Question

An aeroplane takes off from a runway of length 800m. If it leaves the ground in 16 sec. Find its acceleration and the velocity of take off.

Answer 1

U=0m/s

V=?

S=800m

T=16s

A=?

S=ut+1/2 at^2

800=0*16+1/2 *a*16*16

800=0+128a

A=800/128=6.25m/s^2

Therefore, acceleration =6.25m/s^2

V=u+at

V=0+6.25*16

Therefore, v=100m/s

Answer 2

We use the equations of motion

$s= ut+ \frac{1}{2} at^2$   and  $v=u+at$

We are given that the runway is of length 800m and it takes 16 sec to leave the ground, so for a plane at rest initially we decipher,

$s=800m\\t=16s\\u=0$

therefore the equation $s= ut+ \frac{1}{2} at^2$ becomes

$800=0(16)+\frac{1}{2} a(16)^2\\ = \frac{1}{2}a(256)\\ a=\frac{1600}{256} \\a=6.25m/s^2$

Now we know the  accelereation. To find the velocity at the time of take off we use the second equation

$v=u+at\\v=0+(6.25)(16)\\v=100m/s$

Therefore the velocity at the time of take off is 100 m/s and the acceleration is 6.25$m/s^2$.