Question

1. A consumer magazine tested two kinds of engines. One was a standard

engine, and it was determined that its acceleration could be modeled by

f (t) = 6 + .7t feet per square second, t seconds after starting from rest. The acceleration of the turbo-charged model could be approximated by

g(t) = 6 + 6.1 t + .05 t^2 feet per square second, t seconds after starting

from rest. How much faster is the turbo-charged model moving than the standard model at the end of a 10 second trial?

Answer 1

Answer:

$\Delta v = 287\,\frac{ft}{s}$

Step-by-step explanation:

First, velocity function are found by means of integration, knowing that both engines start at rest and, lastly, velocities are evaluated at given time:

Standard Engine

$v_{f} = 6\cdot t +0.35\cdot t^{2}$

$v_{f} (10) = 6\cdot (10) + 0.35\cdot (10)^{2}$

$v_{f} (10) = 95\,\frac{ft}{s}$

Turbocharged Engine

$v_{g} = 6\cdot t + 3.05\cdot t^{2} + 0.017\cdot t^{3}$

$v_{g} (10) = 6\cdot (10) + 3.05\cdot (10)^{2} + 0.017\cdot (10)^{3}$

$v_{g} (10) = 382\,\frac{ft}{s}$

Finally, the difference of the velocity of the turbocharged model with respect to the standard one is:

$\Delta v = 382\,\frac{ft}{s} - 95\,\frac{ft}{s}$

$\Delta v = 287\,\frac{ft}{s}$