problem solving

A roller coaster at an amusement park starts out on a level track 50.0 m long and then goes up a 25.0-m incline at an angle of 30o to the horizontal. It then goes down a 15.0-m ramp with an incline of 40o to the horizontal. When the roller coaster has reached the bottom of the ramp, what is its displacement from the starting point?

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Quote:

Originally Posted by tacubo

A roller coaster at an amusement park starts out on a level track 50.0 m long and then goes up a 25.0-m incline at an angle of 30o to the horizontal. It then goes down a 15.0-m ramp with an incline of 40o to the horizontal. When the roller coaster has reached the bottom of the ramp, what is its displacement from the starting point?

Hello,

with your problem you have to calculate two different movements:

1. vertical direction. Use the Sine function to calculate the height h (red line):

$h=25m\cdot \sin(30^{\circ})+15m\cdot \sin(-40^{\circ})$ $\approx 12.5m-9.64m\approx 2.86m$

2. horizontal direction. Use the Cosine function to calculate the distance d:

$d=50m+25m\cdot\cos(30^{\circ})+15m\cdot\cos(-40^{\circ})$ $\approx50m+21.65m+11.49m\approx83.14m$

3. If you mean by displacement the distance (D) between the starting point and the endpoint of the trip, you have to use the Pythagorian Theorem (blue line):

$D\approx\sqrt{(83.14m)^2+(2.86m)^2}\approx 83.19m$

Greetings

EB

Hello, tacubo!

I assume you made a sketch . . .

Quote:

A roller coaster at an amusement park starts out on a level track 50.0 m long
and then goes up a 25.0-m incline at an angle of 30° to the horizontal.
It then goes down a 15.0-m ramp with an incline of 40° to the horizontal.
When the roller coaster has reached the bottom of the ramp,
what is its displacement from the starting point?

Code:

C Q * - - - - + * : * 40° : 25* : * : * : 15* : * : * : * 30° : * * - - - - - - - - * - - - - - + - - - - D A 50 B P

The roller coaster starts at $A$ , goes 50 m to $B.$

Then it goes up a 25-m incline at 30°.
In right triangle $CPB:\;CP = 25\sin30^o = 12.5,\;\;BP = 25\cos30^o = 21.65$

Then it goes down a 15-m incline at 40°.
In right triangle $CQD:\;QD = 15\sin40^o = 9.64,\;\;CQ = 15\cos40^o = 11.49$

Here is the diagram with the distances labelled.

Code:

C 11.49 Q * - - - - + * : * : * : * :9.64 * : * : * :12.5 * : * : * * - - - - - - - - * - - - - - + - - - - D A 50 B 21.65 P 11.49

Can you finish the problem now?