Arunaway truck with failed brakes is moving downgrade at 130 km/h just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of theta = 15 the truck's mass is 1.2 x 10⁴ kg. (a) what minimum length l must the ramp have if the truck is to stop momentarily) along it? (assume the truck is particle, and justify that assumption.) does the minimum length l rncrease, decrease, or remain the same if (b) the truck's mass is decreased and (c) its speed is decreased

  L = 308.3 m

Explanation:

a) Let's use energy conservation to solve this exercise, as there is no rubbing of the energy in the colon

Initial

        Em₀ = K = ½ m v²

Final

        E = U = mg y

       Em₀ = Emf

       ½ m v² = m g y

       y = ½ v² / g

Let's reduce to SI units

     y = 130 km / h (1000m / 1km) 1 h / 3600s) = 36.1 m / s

     y = ½  1.2 10⁴ 36.1² / 9.8

     y = 79.79 m

This is the vertical height, let's use trigonometry to find the length of the ramp

     sin 15 = y / L

     L = y / sin 15

     L = 79.79 / sin 15

     L = 308.3 m

b)  for this analysis we write in an explanation form

     L = ½ v² / (g sin15)

If truck mass decreases. We can see that the mass does not appear in the formula so the length should not vary

If the velocity decreases, we see that the velocity is in quadratic form in the formula, therefore, as the distance decreases it must decrease in the form of square root

the bal will not change its velocity.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y