Question

Uphill escape ramps are sometimes provided to the side of steep downhill highways for trucks with overheated brakes. for a simple 14 â upward ramp, what minimum length would be needed for a runaway truck traveling 130 km/h ?

Answer 1

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

$a = -g\sin\theta$

$a=-9.8\times\sin14^{\circ}$

$a=-2.37\ m/s^2$

We need to calculate the distance

Using equation of motion

$v^2-u^2=2as$

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

Put the value in the equation

$s=\dfrac{0-(36.11)^2}{2\times(-2.37)}$

$s=275.09\ m$

Hence, The minimum length is 275.09 m

Answer 2

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

$a = -g sin\theta$

$a = -g sin14 = -9.8 sin14$

$a = -2.4 m/s^2$

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

$v_i = 130 * \frac{5}{18}$

$v_i = 36.11 m/s$

now we can use kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 36.11^2 = 2*(-2.4)*d$

$d = 271.7 m$

so the length of inclined plane will be 271.7 m