Find the magnitude of the total displacement, in meters, of the rollercoaster

Question

A roller coaster first travels down a straight track at a distance of 42.5 m at an angle of 55.0^O below the horizontal. It then continues along a straight horizontal section of the track for a distance of 26.0 m. Finally, it travels up a straight track a distance of 38.7 m at an angle of 18.0^O above the horizontal.

Answer 1

You are looking for the red vector in the diagram:

Displacement vector #1 (d₁) can be considered to be straight over (positive x-direction) plus straight down (negative y-direction). These are the "components" of d₁. To find out what they are, use trig ratios:

cos(55°) = d_1-x/42.5 or d_1-x = 42.5cos(55°) = 24.377 m

sin(55°) = d_1-y/42.5 or d_1-y = 42.5sin(55°) = -34.814 m (we had to add the negative manually to account for the down direction).

Displacement vector d₂ is broken down into components d_2-x = 26 m and d_2-y = 0 m

Displacement vector d₃ is broken down into components d_3-x = 38.7cos(18°) = 36.806 m and d_3-y = 38.7sin(18°) = 11.959 m

The total displacement vector (d_T) in the x-direction is the sum of d_1-x, d_2-x, and d_3-x:

d_T-x = 24.377 + 26 + 36.806 = 87.183 m

d_T-y = -34.814 + 0 + 11.959 = -22.855 m

To combine these back together, first get the magnitude by using the Pythagorean Theorem:

d_T = √(87.183² + 22.855²) = 90.1 m

To get the angle, use inverse tangent: θ = tan^-1(22.855/87.183) = 14.7° below the horizon