Problem 2

An athlete swims the length of a 50.0-m pool in 20.0-s and makes the return trip to the starting position in 22.0-s. Determine her average velocities in the following cases:

(a) In the first half of the swim.

SOLUTION:

                                    Remember: average velocity is the total displacement divided by the total elapsed time. The displacement in the first half is 50.0-m since the initial position is 0 and the final position is 50.0-m.

                                    Average velocity = (50.0 m)/(20.0 s) = 2.50 m/s

                                    Note: it makes sense that it is positive because if we define moving to the right, away from the origin or start position as positive, then we certainly expect her displacement to be positive during this half of the swim.

(b) In the second half of the swim.

SOLUTION:

                                    During the second half, the initial position is 50.0-m and the final position is 0.Therefore the displacement is equal to –50.0 m.

                                    Average velocity = (-50.0 m)/(22.0 s) = -2.27 m/s

                                    Again, it makes perfect sense to have a negative velocity since we are moving toward the origin or starting point and we defined our negative direction to be toward the left.

(c) The entire swim.

SOLUTION:

                                    Without even solving we can say that the answer is ZERO. This is because she returned to her original position hence her initial and final positions are 0.

 

Problem 4

A certain bacterium swims with a speed of 3.5 mm/s. How long will it take this bacterium to swim across a petri dish having a diameter of 8.4 cm?

SOLUTION:

Let us first convert to each quantity into SI units:

                                    3.5 mm/s = 3.5 x 10-6 m/s

                                    8.4 cm = 8.4 x 10-2 m

Now we can solve for time using: v = d/t, therefore t = d/v

                                    t = d/v = (8.4 x 10-2 m)/ (3.5 x 10-6 m/s) = 2.4 x 104 s.

 

Problem 6

A tennis player moves in a straight-line path as shown in the figure below. Find her average velocities in the following time intervals.

(a) 0 to 1.0 s

SOLUTION:

                                    x(initial) = 0      and       t(initial) = 0

                                    x(final) = 4 m    and       t(final) = 1.0 s

                                    Dx = 4 m – 0 = 4 m

                                    Dt = 1.0 s- 0 = 1.0 s

                                    v(average) = (4 m)/(1.0 s) = 4 m/s

(b) 0 to 4.0 s

SOLUTION:

                                    x(initial) = 0      and       t(initial) = 0

                                    x(final) = -2m   and       t(final) = 4.0 s

                                    Dx = -2 m – 0 = -2 m

                                    Dt = 4.0 s- 0 = 4.0 s

                                    v(average) = (-2 m)/(4.0 s) = -0.5 m/s

(c) 1 to 5.0 s

SOLUTION:

                                    x(initial) = 4 m and        t(initial) = 1.0 s

                                    x(final) = 0        and       t(final) = 5.0 s

                                    Dx = 0 – 4 m = -4 m

                                    Dt = 5.0 s- 1.0 s = 4.0 s

                                    v(average) = (-4 m)/(4.0 s) = -1 m/s

(d) 0 to 5.0 s

SOLUTION:

                                    x(initial) = 0 m and        t(initial) = 0

                                    x(final) = 0        and       t(final) = 5.0 s

                                    Dx = 0 – 0 = 0

                                    Dt = 5.0 s- 0 = 5.0 s

                                    v(average) =(0)/(5.0 s) = 0

 

Problem 8

If the average speed of the orbiting space shuttle is 19,800 mi/h, determine the time required for it to circle the Earth. Make sure you consider the fact that the shuttle is orbiting about 200 mi above the surface of the Earth, and assume the Earth’s radius is 3963 miles.

SOLUTION:

Assume that the shuttle has a circular orbit around the Earth. To find the total distance traveled in one orbit or to circle the Earth, you must realize that you must find the circumference of the circular path traced out by the shuttle. Recall that the circumference of a circle is C = 2pR, where R is the radius of the circle. What is this radius?

This radius is measured from the center of the Earth, therefore:

                                    R = 3963 mi + 200 mi = 4163 mi.

                                    C = distance = 2p(4163) = 26157

                                    V = distance/time 

                                    Time = distance/speed = (26157 mi)/(19,800 mi/h) = 1.32 h.

 

Problem 12

Runner A is initially 4.0 mi west of a flagpole and is running with a constant velocity of 6.0 mi/h due East. Runner B is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far from the flagpole are the runners when they meet?

SOLUTION:

This is a more challenging problem. We start out by identifying our coordinate axis and origin. We take the flagpole to be the origin and the East direction as positive and West direction as negative. Each runner has an equation of motion written relative to the location of the flagpole:

RUNNER A: he is 4.0 mi west of the pole and running east. His initial location is negative, and his velocity is positive:

                                    X(A) = -4.0 mi + (6.0 mi/h)t

RUNNER B: he is 3.0 mi east of the flagpole and running west. His initial location is positive and his velocity is negative:

                                    X(B) = 3.0 mi - (5.0 mi/h)t

We first find the time it takes for them to meet by setting each equation equal to each other then solve for t:

                                    X(A) = X(B)

                                    -4.0 mi + (6.0 mi/h)t = 3.0 mi - (5.0 mi/h)t

                                    (11.0 mi/h)t = 7.0 mi

                                    t = (7.0 mi)/(11.0 mi/h) = 0.636 h

Now that we know when they meet, we can find their distance to the flagpole by plugging this time into either equation. To check your answer, plug it into both equations and you SHOULD get the same answer:

                                    X(A) = -4.0 mi + (6.0 mi/h)(0.636 h) = -0.18 m

                                    X(B) = 3.0 mi – (5.0 mi/h)(0.636 h) = -0.18 m.

Voila, the answers match which means we must have done something right!!