# Need Help! - Equilateral Triangle

## Homework Statement

A particle moves from rest and has impressed a uniform velocity of 10 meters per second parallel to one side of an equilateral triangle, and a uniform acceleration of 10 meters/sec2 parallel to an adjacent side of the triangle. Find the distance of the particle from its original position at the end of 5 seconds.

## The Attempt at a Solution

Can someone walk me through this? Been working on it for a couple days and can't figure it out.

## Homework Statement

A particle moves from rest and has impressed a uniform velocity of 10 meters per second parallel to one side of an equilateral triangle, and a uniform acceleration of 10 meters/sec2 parallel to an adjacent side of the triangle. Find the distance of the particle from its original position at the end of 5 seconds.

## The Attempt at a Solution

Can someone walk me through this? Been working on it for a couple days and can't figure it out.

No walking but I take you have done 2-D projectile motion.

Well pretend like a horizonal side of the triangle is the x-axis and the motion does not involve acceleration. Now pretend like gravity is acting up (I assume that was +10 m/s/s) and at a 60 degree angle (equilateral triangle) with respect to the x-axis, instead of perpendicular and down like in projectile motion...

No walking but I take you have done 2-D projectile motion.

Well pretend like a horizonal side of the triangle is the x-axis and the motion does not involve acceleration. Now pretend like gravity is acting up (I assume that was +10 m/s/s) and at a 60 degree angle (equilateral triangle) with respect to the x-axis, instead of perpendicular and down like in projectile motion...

have only done very little 2-d projectile motion. this is my first physics class and i feel a little in over my head!

would this be a horizontal projectile problem?
-vertical component of inital velocity is 0 = 10sin60 = 8.66
-horizontal component = 10cos60 = 5

Then find the horizontal displacement:

To find the horizontal displacement at 4.0 s :

d = vit + (0.5)at2 = (10)(5) + (0.5)(10)(5)2 = 175 meters??

have only done very little 2-d projectile motion. this is my first physics class and i feel a little in over my head!

would this be a horizontal projectile problem?
-vertical component of inital velocity is 0 = 10sin60 = 8.66
-horizontal component = 10cos60 = 5

Then find the horizontal displacement:

To find the horizontal displacement at 4.0 s :

d = vit + (0.5)at2 = (10)(5) + (0.5)(10)(5)2 = 175 meters??

There is no vertical component of a projectile launched horizontally... if that is what you are asking...
The initial problem is a trumped up version of a projectile problem. I dont mean to be rude, but either this is extra credit, or you might have missed the boat on projectile motion or motion in 2-D and your teacher is moving on thinking you got it down.

In the first problem you will have to find out how far an object moving at a constant velocity(10 m/s) will travel in a given time in the x- direction (to make it easy). Then you are going to find out how far that same particle moves at an angle of 60 degrees with respect to the horizontal while accelerating at 10 m/s/s. Once you have found both of these displacement vectors add them together.

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