Question

The position of a particle along a straight-line path is defined by s=(t3−6t2−15t+7) ft, where t is in seconds. part a part complete determine the total distance traveled when t = 11 s . express your answer to three significant figures and include the appropriate units. st = 640 ft previous answers correct part b what are the particle's average velocity at the time given in part a? express your answer to three significant figures and include the appropriate units.

Answer 1

A) The position of the particle along the straight-line path is given by
$s=(t^3-6t^2-15 t+7)ft$
where t is the time.
If we substitute t=11 s inside the equation, we find the total distance covered during this time, in feet:
$s=(11)^3-6(11)^2-15(11)+7=597 ft$
which converted into meters is s=182 m.

b) The average velocity of the particle at t=11 s is equal to the total distance travelled, s=597 ft, divided by the time taken, t=11 s:
$v= \frac{s}{t}= \frac{597 ft}{11s} =54.3 ft/s$
We can also calculate it in meters/second:
$v= \frac{s}{t}= \frac{182 m}{11 s}=16.5 m/s$