Uphill escape is designed to provide a safe braking system
In this type of uphill system when truck use it, the truck will be stopped due to action of gravity
In this system we can say the acceleration due to gravity along the inclined is given as
[tex]a = -g sin\theta[/tex]
[tex]a = -g sin14 = -9.8 sin14[/tex]
[tex]a = -2.4 m/s^2[/tex]
now due to this acceleration we can say the truck will stop after traveling distance "d"
now the initial speed of the truck is 130 km/h
[tex]v_i = 130 * \frac{5}{18}[/tex]
[tex]v_i = 36.11 m/s[/tex]
now we can use kinematics
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 36.11^2 = 2*(-2.4)*d[/tex]
[tex]d = 271.7 m[/tex]
so the length of inclined plane will be 271.7 m
Answer:
The minimum length is 275.09 m
Explanation:
Given that,
Angle = 14°
Speed = 130 km/h = 36.11 m/s
We need to calculate the distance
According to figure,
We need to calculate the deceleration
[tex]a = -g\sin\theta[/tex]
[tex]a=-9.8\times\sin14^{\circ}[/tex]
[tex]a=-2.37\ m/s^2[/tex]
We need to calculate the distance
Using equation of motion
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
Put the value in the equation
[tex]s=\dfrac{0-(36.11)^2}{2\times(-2.37)}[/tex]
[tex]s=275.09\ m[/tex]
Hence, The minimum length is 275.09 m
