A physics instructor drives down to St. Louis to see his mom. On the way there,
ID: 1625834 • Letter: A
Question
A physics instructor drives down to St. Louis to see his mom. On the way there, he travels the first half of the time at 60 km/hr and the second half the time at 95 km/hr. Coming back, he drives the first half the distance at 60 km/hr and the second half of the distance at 95 km/hr. What is his average speed going down to St. Louis? S_avg = What is his average speed coming back from St. Louis? What is his average speed during the entire trip? S_avg = What is his total displacement during the entire trip? Delta X = Which of these above graphs best represents his velocity during the trip to St. Louis? Plot = Put in the number below the correct graph Which of these above graphs best represents his displacement during the trip to St. Louis? Plot = Put in the number below the correct graphExplanation / Answer
1)Average speed=Path length/Time interval
let T be the time interval and distance travelled be x.
Average speed= (60×T/2)+(95×T/2)/T
=(60+95)/2
=77.5 km/hr
2)Average speed=path length/time taken
let the total path length be 2x
time taken=path length/speed
Average speed=2x/(x÷60)+(x÷95)
=2×60×95/(60+95)
=73.548 km/hr.
3)Average speed during the entire trip= (77.5×T)+(73.548×T)/2T
=75.524 km/hr.
4) The magnitude of displacement for a course of motion may be zero,unlike path length.Therefore the total displacement is zero for the entire trip.
5)The graph 3 represents his velocity ,since the first half of time he travels with a constant velocity of 60km/hr and second half of time with 95 km/hr ,since the velocity is constant we obtain a straight line parallel to time axis.
6)The graph 2 represents the displacement,that is he covers equal distances in equal intervals of time therefore the motion is uniform and obtains a straight line graph.
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