Question

To get to a concert in time, a harpsichordist has to drive 122 miles in 1.93 hours. (a) If he drove at an average speed of 52.0 mi/h in a due west direction for the first 1.22 h, what must be his average speed if he is heading 30.0° south of west for the remaining 42.6 min?(b) What is his average velocity for the entire trip?(give magnitude and direction)

Answer 1

The key difference between speed and velocity is that velocity includes a direction.

(a) Let x be the unknown speed during the second leg of his trip. We know that d = rt, and we are given that the total trip is 122 miles and 1.93 hours, so

52 mph * 1.22 hrs + x * (1.93 hrs - 1.22 hrs) = 122 miles

Solve for x.

==> x = 82.47887 mph

(b) Average velocity is the displacement (i.e. straight line distance of the final position from the original position combine with a direction) divided by the elapsed time.

Let due west be direction 0 degrees and positive angles are measures counterclockwise (i.e. South of West).

During the first leg of the trip, his displacement was due West. The distance is (52mph * 1.22 hrs) = 63.44 miles.

The length of the second leg is 122 miles minus the length of the first leg (63.44) which is 58.56. We will decompose this into a due West part and a due South part.

dw = 58.56 * cos(30)

ds = 58.56 * sin(30)

The total displacement is found by adding the displacements for the two legs.

Dw = 63.44 + (58.56 * cos(30)) = 114.15

Ds = 0.00 + (58.56 * sin(30)) = 29.28 { the first leg didn't have an southerly component }

We can convert the westerly and southerly components back into a distance and direction by using

D = sqrt((Dw)^2 + (Ds)^2) = 117.845

degrees South of West = arctan(Ds / Dw) = 14.386

The average velocity vector will have the same direction as the displacement vector.

The magnitude will be D / 1.930 = 61.0595 mph

Direction is 14.386 degrees south of west