Question

An aeroplane after completing journey lands on the airstrip at a velocity of 100 m/s. After 40 second it stops. How much distance it will cover during this time?​

Answer 1

Answer:-

• initial velocity=u=100m/s
• Time=t=40s
• Final velocity=v=0m/s
• Distance=s=?

Finding acceleration:-

$\boxed{\sf Acceleration=\dfrac{v-u}{t}}$

$\\ \rm\longmapsto Acceleration=\dfrac{0-100}{40}$

$\\ \rm\longmapsto Acceleration=\dfrac{-100}{40}$

$\\ \rm\longmapsto Acceleration=-2.5m/s^2$

According to 2nd equation if motion

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

$\\ \rm\longmapsto s=100(40)+\dfrac{1}{2}\times (-2.5)(40)^2$

$\\ \rm\longmapsto s=400+(-2.5)\times 400$

$\\ \rm\longmapsto s=400+(-1000)$

$\\ \rm\longmapsto s=400-1000$

$\\ \rm\longmapsto s=-600$

• Distance must be positive

$\\ \rm\longmapsto s=600m$