Given that,
Deacelectration = -15 m/s²
Negative sign shows the declaration
Slowly speed = 25 m/s
Acceleration = 35 m/s²
Speed = 170 m/s
We need to calculate the time
Using equation of motion
[tex]v=u+at[/tex]
Where, v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the formula
[tex]25=0+15\times t[/tex]
[tex]t=\dfrac{25}{15}[/tex]
[tex]t=1.66\ sec[/tex]
We need to calculate the distance from the original runway
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
[tex]s=0+\dfrac{1}{2}\times35\times(1.66)^2[/tex]
[tex]s=48.2\ m[/tex]
Hence, The distance from the original runway is 48.2 m.
