Answer: 19 meters.
Explanation:
We want to find the total displacement between t = 2s and t = 4s.
To do it, we can integrate our function, first write our velocity equation.
for t ≤ 3s, we have a linear equation, let's write it:
A linear relationship can be written as:
y = a*t + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
Now we can see that our line passes through the points (1, 0) and (0, -2)
then the slope is:
a = (0 -(-2)/(1 - 0) = 2/1 = 2
and knowing that when t = 0s, v(0s) = -2m/s, then our equation is:
v(t) = (2m/s^2)*t - 2m/s for t ≤ 3s
now, for t ≥3s the equation is constant, v(t) = 4m/s.
then we have
v(t) = (2m/s^2)*t - 2m/s -------if t ≤ 3s
v(t) = 4m/s ----- if t ≥ 3s
Now we integrate over time to get the position:
for t ≤ 3s we have:
p(t) = (1/2)*(2m/s^2)*t^2 - 2m/s*t + C
where C is a constant of integration, as we are calculating the displacement this constant actually does not matter, so we can use C = 0m
p(t) = (1m/s^2)*t^2 - 2m/s*t for t ≤ 3s
and p(3s) = (1m/s^2)*3s^2 - 2m/s*3s = 9m - 6m = 3m is the initial position of the other part of the function.
for t ≥ 3s we have:
p(t) = 4m/s*t + p(3s) = 4m/s*t + 3m
then the position equation is:
p(t) = (1m/s^2)*t^2 - 2m/s*t ---- t ≤ 3s
p(t) = 4m/s*t + 3m --- if t ≥ 3s
Now the displacement will be:
p(4s) - p(2s) where for each time, you need to use the correct function:
p(4s) = 4m/s*4s + 3m = 16m + 3m = 19m
p(2s) = (1m/s^2)*2s^2 - 2m/s*2s = 4m - 4m = 0m
p(4s) - p(2s) = 19m - 0m = 19m
