Answer to Question #262573 in Calculus for ahmed

Question

The power from an engine (in W) is represented by Power, where t is the time in seconds.

1. Draw a graph of power against time
2. Show the area on the graph which represents the energy converted between 5 and 10 seconds
3. Evaluate (using the rules of integration) the energy between 5 and 10 seconds

Part of the machine moves with a velocity given by the expression mm/s

1. Calculate the displacement over the first 5 seconds of motion.

Answer 1

given $p=120t^{1.8}+8t$

Part A.

The graph of the function is shown below.

Part B.

The area corresponding to the energy between 5 and 10 second is shown below (in green)

Part C

The energy corresponding to the area under the graph is found from.

$E=\int^{10}_{5} p(t)dt\\=\int^{10}_{5}(120t^{1.8}+8t)dt\\=[\frac{120}{2.8}t^{1.8}+\frac{8t^2}{0}]^{10}_{5}\\=23458.28J$

Part D

First find the displacement function by integrating the velocity function and evaluting is at t=5 seconds

$d=\int^{t}_{0}(10^2+0.6x)dx\\=[\frac{10}{3}x^3+\frac{0.6x^2}{2}]^{t}_{0}\\\frac{10t^3}{3}+0.3t^2$

or

$d(5)-\frac{10}{3}.5^3+0.3.5^2-424.17mm$