In this equation, the variables are: Displacement = s . The distance the object has traveled from its starting position. [3] X Research source For example, if an object goes 10 meters forward and 7 meters backward, its total displacement is 10 - 7 = 3 meters (not 10 + 7 = 17 meters). Time = t . Self explanatory. Typically measured in seconds.
In this equation, the variables are: Displacement = s . The distance the object has traveled from its starting position. [3] X Research source For example, if an object goes 10 meters forward and 7 meters backward, its total displacement is 10 - 7 = 3 meters (not 10 + 7 = 17 meters). Time = t . Self explanatory. Typically measured in seconds.
In this equation, the variables are: Displacement = s . The distance the object has traveled from its starting position. [3] X Research source For example, if an object goes 10 meters forward and 7 meters backward, its total displacement is 10 - 7 = 3 meters (not 10 + 7 = 17 meters). Time = t . Self explanatory. Typically measured in seconds.
In other words, start by going through the “t” side of your equation from left to right. Every time you reach a “t”, subtract 1 from the exponent and multiply the entire term by the original exponent. Any constant terms (terms which don’t contain “t”) will disappear because they be multiplied by 0. This process isn’t actually as hard as it sounds — let’s derive the equation in the step above as an example:s = -1. 5t2 + 10t + 4(2)-1. 5t(2-1) + (1)10t1 - 1 + (0)4t0-3t1 + 10t0-3t + 10
In other words, start by going through the “t” side of your equation from left to right. Every time you reach a “t”, subtract 1 from the exponent and multiply the entire term by the original exponent. Any constant terms (terms which don’t contain “t”) will disappear because they be multiplied by 0. This process isn’t actually as hard as it sounds — let’s derive the equation in the step above as an example:s = -1. 5t2 + 10t + 4(2)-1. 5t(2-1) + (1)10t1 - 1 + (0)4t0-3t1 + 10t0-3t + 10
In our running example, our finished equation should now look like this:ds/dt = -3t + 10
Note that we use the label “meters/second” above. Since we’re dealing with displacement in terms of meters and time in terms of seconds and velocity in general is just displacement over time, this label is appropriate.
To graph an object’s displacement, use the x axis to represent time and the y axis to represent displacement. Then, just plot points by plugging values for t into your displacement equation, getting s values for your answers, and marking the t,s (x,y) points on the graph. Note that the graph can extend below the x axis. If the line representing your object’s motion drops below the x axis, this represents your object moving behind where it started. Generally, your graph won’t extend behind the y axis - we don’t often measure velocity for objects moving backward in time!
Let’s say that our displacement line contains the points (1,3) and (4,7). In this case, if we want to find the slope at (1,3), we can set (1,3) = P and (4,7) = Q.
In our example, as we moved Q closer to P, we got values of 1. 8, 1. 9, and 1. 96 for H. Since these numbers appear to be approaching 2, we can say that 2 is a good estimate for the slope at P. Remember that the slope at a given point on a line is equal to the derivative of the line’s equation at that point. Since our line is showing our object’s displacement over time and, as we saw in the section above, an object’s instantaneous velocity is the derivative of its displacement at a given point, we can also say that 2 meters/second is a good estimate for the instantaneous velocity at t = 1.
First, we’ll take our equation’s derivative:s = 5t3 - 3t2 + 2t + 9s = (3)5t(3 - 1) - (2)3t(2 - 1) + (1)2t(1 - 1) + (0)9t0 - 115t(2) - 6t(1) + 2t(0)15t(2) - 6t + 2 Then, we’ll plug in our value for t (4):s = 15t(2) - 6t + 215(4)(2) - 6(4) + 215(16) - 6(4) + 2240 - 24 + 2 = 218 meters/second
First, let’s find Q points at t = 2, 1. 5, 1. 1 and 1. 01. s = 4t2 - tt = 2: s = 4(2)2 - (2)4(4) - 2 = 16 - 2 = 14, so Q = (2,14)t = 1. 5: s = 4(1. 5)2 - (1. 5)4(2. 25) - 1. 5 = 9 - 1. 5 = 7. 5, so Q = (1. 5,7. 5)t = 1. 1: s = 4(1. 1)2 - (1. 1)4(1. 21) - 1. 1 = 4. 84 - 1. 1 = 3. 74, so Q = (1. 1,3. 74)t = 1. 01: s = 4(1. 01)2 - (1. 01)4(1. 0201) - 1. 01 = 4. 0804 - 1. 01 = 3. 0704, so Q = (1. 01,3. 0704) Next, let’s get our H values:Q = (2,14): H = (14 - 3)/(2 - 1)H = (11)/(1) = 11Q = (1. 5,7. 5): H = (7. 5 - 3)/(1. 5 - 1)H = (4. 5)/(. 5) = 9Q = (1. 1,3. 74): H = (3. 74 - 3)/(1. 1 - 1)H = (. 74)/(.
- = 7. 3Q = (1. 01,3. 0704): H = (3. 0704 - 3)/(1. 01 - 1)H = (. 0704)/(.
- = 7. 04 Since our H values seem to be getting very close to 7, we can say that 7 meters/second is a good estimate for the instantaneous velocity at (1,3).
