Question

The power (in watts) from an engine is given by the equation P=〖80t〗^1.3+5t where t is the time in seconds.

Draw a graph of power against time for the engine. From the graph approximate the energy produced between 3 and 7 seconds.
Using an appropriate method of integration to calculate the energy produced, and compare the answer with your approximation.

Answer 1

Answer on Question # 59213 – Physics – Mechanics | Relativity

The power (in watts) from an engine is given by the equation $\mathrm{P} = (80\mathrm{t})^{1.3} + 5\mathrm{t}$ where $t$ is the time in seconds. Draw a graph of power against time for the engine. From the graph approximate the energy produced between 3 and 7 seconds. Using an appropriate method of integration to calculate the energy produced, and compare the answer with your approximation.

Solution:

The graph of power against time (Figure 1):

Figure 1 - Power against time

The energy produced between 3 and 7 seconds is numerically equal to the area under the graph, enclosed by the graph, the x axis and vertical lines $t = 3$ s and $t = 7$ s. It can be approximately calculated as the total area of one rectangle and one triangle:

$$\begin{array}{l} \mathrm {E} _ {\text {a r e a}} = \mathrm {S} \left[ (3; 0), (3; \mathrm {P} (3)), (7; \mathrm {P} (7)), (7; 0) \right] = \\ \approx S _ {\text {r e c t}} \left[ (3; 0), (3; P (3)), (7; P (3)), (7; 0) \right] + S _ {\text {t r i}} \left[ (3; P (3)), (7; P (7)), (7; P (3)) \right] = \\ = (7 - 3) \times (\mathrm {P} (3) - 0) + \frac {(7 - 3) \times (\mathrm {P} (7) - \mathrm {P} (3))}{2} = \\ = (7 - 3) \times (\mathrm {P} (3) - 0) + \frac {(7 - 3) \times (\mathrm {P} (7) - \mathrm {P} (3))}{2} = \\ = (7 - 3) \times ((8 0 \times 3) ^ {1. 3} + 5 \times 3) + \frac {(7 - 3) \times (\left[ (8 0 \times 7) ^ {1 . 3} + 5 \times 7 \right] - \left[ (8 0 \times 3) ^ {1 . 3} + 5 \times 3 \right])}{2} = \\ \end{array}$$ $$= (7 - 3) \times ((80 \times 3)^{1.3} + 5 \times 3) + \frac{(7 - 3) \times ([ (80 \times 7)^{1.3} + 5 \times 7] - [ (80 \times 3)^{1.3} + 5 \times 3])}{2} = 10060.96\ [\mathrm{J}].$$

The energy can also be calculated by integrating the equation of power over the time:

$$\begin{array}{l} E_{\text{int}} = \int_{3}^{7} ((80t)^{1.3} + 5t) \, dt = \left(80^{1.3} \frac{t^{2.3}}{2.3} + 5 \frac{t^{2}}{2}\right) \int_{3}^{7} = \\ = \left(80^{1.3} \times \frac{7^{2.3}}{2.3} + 5 \times \frac{7^{2}}{2}\right) - \left(80^{1.3} \times \frac{3^{2.3}}{2.3} + 5 \times \frac{3^{2}}{2}\right) = 11499.16 - 1643.07 = 9856.09\ [\mathrm{J}]. \end{array}$$

The relative deviation of the approximate value of energy from the exact value:

$$\delta = \frac{\left| E_{\text{int}} - E_{\text{area}} \right|}{E_{\text{int}}} \times 100\% = \frac{\left| 9856.09 - 10060.96 \right|}{9856.09} \times 100\% = 2.08\ [\%].$$

Since the graph of power against time is close to linear, the relative deviation of the approximate value of energy from the exact value is small.

Answer: $E_{\text{area}} \approx 10060.96\ [\mathrm{J}]$; $E_{\text{int}} = 9856.09\ [\mathrm{J}]$; $\delta = 2.08\ [\%]$.

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