Question

An airbus runs on a runway from rest. Its final velocity is 810Km/hr. The length of the runway is 1000m. Calculate the uniform acceleration of the air bus.

Answer 1

Given :

Initial velocity of the airbus : 0 m/s

Final velocity of the airbus : 810 km/hr

Distance : 1000 m

To find :

The uniform acceleration of the airbus.

Solution :

Firstly we have to convert km/hr to m/s

we know that,

» 1km/hr = 5/18km/hr

» 810km/hr = 810 × 5/18 = 225 m/s

Accerlation is the rate of change of velocity.

• It is a vector quantity having both magnitude as well as direction.
• SI unit of acceleration is metre per second square.

As we are provided with initial velocity, final velocity and distance we can use second equation of kinematics that is,

v² - u² = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

by substituting all the given values in the formula,

➝ v² - u² = 2as

➝ (810)² - (0)² = 2(a)(1000)

➝ 656100 = 2000a

➝ a = 656100/2000

➝ a = 328.05

Thus, the acceleration of the air bus is 328.05 m/s².

Know more :

First equation of kinematics : v = u + at

Second equation of kinematics : s = ut + 1/2at²

Third equation of kinematics : v² - u² = 2as

Answer 2

Answer:-

• Initial velocity=u=0m/s
• Final velocity=810km/h

Convert it to m/s

$\\ \sf\longmapsto 810\times \dfrac{5}{18}$

$\\ \sf\longmapsto 225m/s$

• Distance=1000m

According to second equation of kinematics

$\boxed{\sf v^2-u^2=2as}$

$\\ \sf\longmapsto a=\dfrac{v^2-u^2}{2s}$

$\\ \sf\longmapsto a=\dfrac{810^2-0^2}{2(1000)}$

$\\ \sf\longmapsto a=\dfrac{656100}{2000}$

$\\ \sf\longmapsto a=328.05m/s$