The coordinates (0,0) mark the take-off point for a rocket constructed as part of a science class. The positive x direction from (0,0) is considered to be East.
a) Find the equation of the rocket's path if it rises at a rate of 5m vertically for every 1m in an easterly direction.
b) A second rocket is fired 2m vertically above from where the first rocket was launched. It rises at a rate of 13m for every 2m in an easterly direction. Find the equation describing its path.
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Here, it's best to understand the idea of slope. It is defined as the ratio of the change in the vertical dimension over the corresponding change in the horizontal dimension. As an equation, it looks like this if given two points on the line `(x_1,y_1)` and `(x_2,y_2)` :
`m = (Deltay)/(Deltax) = (y_2-y_1)/(x_2-x_1)`
However, slope is only one part of the picture. You can have many different lines of the same slope! So, you'll need another parameter. For us, the convenient choice will be the y-intercept, we'll call b, meaning the value of the y coordinate when x=0. Knowing the slope and y-intercept, we can produce the following equation for the line:
`y = mx + b`
Armed with this tool, we can now solve the problem.
For the first part, we are given that for any horizontal change of one unit, the vertical change will be 5 units. We are given in the problem that when x = 0, y = 0, as well, considering the starting point is (0,0). So, we have our slope (`m = 5/1`) and we have our y-intercept: 0.
Our equation is therefore the following:
`y = 5/1 x + 0`
`y = 5x`
For (b), the exact same concept applies. We know that for each horizontal change of 2 units, the vertical change is 13 units, giving us a slope of 13/2. At the starting point (x = 0), we are given that it is fired 2 units from the ground (giving a y-intercept of 2). Therefore, we get the following equation describing the motion of the rocket:
`y = 13/2 x + 2`
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