Question

An areoplane taking off from a field has a run of 500 m. What are its acceleration and take off velocity if it leaves the ground 10 seconds after the start ?

Answer :
acceleration{a} = 10m/s square
velocity{V} = 100m/s

But I need solution.
Most satisfying answer will mark as brainliest.
please answer it fastly.
physics genius.... please help.​

Answer 1

Answer:

$\large a=10 \ m/sec^2$

v = 100 m / sec.

Explanation:

Given :

Distance ( s ) = 500 m

Time ( t ) = 10 seconds.

Initial velocity ( u ) = 0 m / sec

From second equation of motion we have

$\large s=ut+\dfrac{1}{2}at^2$

where s =  distance

t = time

a = acceleration

u = Initial velocity

Put the values in formula

$\large 500=0\times10+\dfrac{1}{2}a10^2\\\\\\\large 500=\dfrac{1}{2}a10^2\\\\\\\large \dfrac{500}{100}=\dfrac{1}{2}a\\\\\\\large \dfrac{1}{2}a=5\\\\\\\large a=10 \ m/sec^2$

From first equation of motion we have

v = u + a t

Put the values here

v = 0 + 10 × 10 m / sec

v = 100 m / sec.

Thus we get answer.