A butterfly is flying around and its velocity(v) as a function of time(t) is given in the graph below where rightwards is the positive velocity direction. What is the butterfly's displacement x from t=2 to 4s? Answer with two significant digits.

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

d. learned preferences i believe is your answer

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s