Question

An Aeroplane taking off from a field , has run of 500m. What is the acceleration if it leaves the ground in 10s from start? Also find the take off velocity.

Answer 1

Howdy!!

your answer is -----

Given, distance s = 500m ,
time t = 10s
since, it is start from rest ,
so, initial velocity u = 0

Now, according to formula

$s = ut + \frac{1}{2} a {t}^{2} \\ = > 500 = 0 \times 10 + \frac{1}{2} \times a \times 100 \\ = > 500 = 50a \\ = > a = \frac{500}{50} = 10m {s}^{ - 2}$

let , take off velocity is v.

so, according to formula

v = u + at
=> v = 0 + 10×10

=> v = 100m/s


hence, acceleration a =10m/s^-2 and take off velocity v = 100m/s


hope it help you

Answer 2

Hey

Given:
DISTANCE = s=500m
Initial velocity =u=0 m/s
Final velocity =V=?
Time=t=10s
From Second equation of motion:
S=ut + 1/2at^2
500=0*10 +1/2 *a 100
a=500/50=10m/s2
From V= u+ at
=0+10*10
=100m/ s
Therefore , acceleration is 10m/ s2 and take off velocity is 100m/s.