2. Relevant equations
P = dW/dt
Change in work = Change in Kinetic Energy
3. The attempt at a solution
Since the integral of the power graph is work done in the system, and since it starts at 0, does this mean kinetic energy is the same thing? So I can probably make a triangle with a height of 20 and base 1 to get an area of 10 for the first problem? And then 30 for the next one?
Though I feel this is much more complicated than that…