Time, Speed, Work and Wages

Man-day work relation

Work in this topic is measured in man-days (or man-hours), where one man-day is the work one person completes in one day. If every person does the same daily amount r, the total work is W = M \times D \times r, where M is the number of men and D the number of days. For a fixed amount of work, the number of men is inversely proportional to the number of days, M \propto \frac{1}{D}, and for a fixed number of men the work is directly proportional to the days. If the number of workers and the amount of work are scaled by the same factor, each worker keeps the same share, so the number of days does not change.

When two situations are compared at the same daily rate, equate or take the ratio of the products M \times D. If a men work b days and c men work d days, and the works are in the ratio p:q, then \frac{ab}{cd}=\frac{p}{q}. When the unknown appears in both the number of men and the number of days, this gives an equation in the unknown, and the root that suits the context is chosen. Equal works give M_1D_1 = M_2D_2 directly.

If the work of a group is given as a fraction of the whole job, convert it to a fraction before forming the equation. For a partial job, the work done by the group in n days plus the work done by the remaining worker in the leftover days must equal the whole work.

When people work together, add their one-day shares, not their times. If one person finishes a job in T days, his one-day share is \frac{1}{T}. The combined one-day share of a group is the sum of these shares, and the time for the whole job is its reciprocal. For example, if the joint rate is \frac{1}{8} of the work per day, then half the work takes \frac{1/2}{1/8} days. Equations built from pairwise times can be solved for individual rates, which are then added.

When a time is given in mixed units such as 1\frac{1}{2} days, first convert it to a fractional rate per day, then add the rates. The total time is the total work divided by the combined rate.

Efficiency is the one-day share of work. Equal work by two groups means the sum of their shares is equal, and the ratio of the shares of one woman and one man can be found from two such equations. Once the individual shares are known, the time for a different group is found by dividing the total work by their joint rate.

If the speed of one worker is k times that of another, the worker's one-day share is k times as large, and the time taken is \frac{1}{k} times as long.

For man-hours, multiply persons, hours per day and days. Keep the total man-hours as the measure of work when the working hours per day change.

When the workforce changes during the job, compute the work done in each period as men multiplied by days, subtract it from the total, and find the new number of men required for the remaining days. Adding men means multiplying the workforce by 1+\frac{x}{100} for an increase of x percent, and the time is then divided by the same factor.

In scheduling problems in which only one person can work in an hour and no person can work in consecutive hours, the rates must be combined in an alternating order to finish the work in the fewest hours.

Speed and time inverse relation

For a fixed distance s, speed and time are related by s = vt, so the speed is inversely proportional to the time: v \propto \frac{1}{t}. For two speeds v_1 and v_2 covering the same distance, v_1 t_1 = v_2 t_2, and the ratio of times is the inverse of the ratio of speeds.

If the speed becomes a fraction \frac{4}{5} of the usual speed, the time becomes \frac{5}{4} of the usual time. The extra time is therefore \frac{1}{4} of the usual time, and this difference can be equated to the given delay to find the usual time.

If the speed doubles, the time for the same distance is halved. When the speed is reduced by a fixed amount, the time difference is \frac{d}{v-k}-\frac{d}{v}, which leads to a quadratic equation in the speed.

When two vehicles cover the same distance with speeds differing by a fixed amount, the difference of their times is \frac{d}{v}-\frac{d}{v+k}=\frac{dk}{v(v+k)}. Keep the time in hours by converting minutes, because 1 hour equals 60 minutes. Solving the resulting relation gives the speed.

When a person is early at one speed and late at another, the difference between the two arrival times equals the sum of the early and late margins. Divide the distance by each speed to write the time difference, and solve for the distance.

For a circular path, the distance for one round is the circumference \pi d (or 2\pi r). Convert the speed to metres per minute before dividing the distance by the speed, since 1 km/hr = \frac{1000}{60} m/min.

Relative speed of two moving objects

When two objects move towards each other, their relative speed is the sum of their speeds, because the gap between them closes at that rate. When they move in the same direction, the relative speed is the difference of their speeds, because the faster one gains on the slower one at that rate.

To find when two objects meet, first find the gap between them at the moment the later one starts, which includes any head start of the earlier one. Divide this gap by the relative speed to get the time after the later start. The meeting time on the clock is this elapsed time added to the later start time, and the meeting point is found by multiplying a speed by the total time of motion of that object.

When a person walks in the same direction as a faster person who starts a distance d behind, the faster person overtakes after time \frac{d}{u-v}, where u and v are the two speeds, and the distance covered by the faster person is \frac{ud}{u-v}.

When a train passes a moving person, the relative speed is the train speed minus the person speed, if they move in the same direction. The length of the train equals this relative speed multiplied by the time taken to pass the person. Speeds in km/hr must be converted to m/s by multiplying by \frac{5}{18} before multiplying by the time in seconds.

When a train crosses a person moving in a train in the opposite direction, the relative speed is the sum of the two train speeds, and the time taken equals the length of the crossing train divided by this sum. In the same direction, the time equals the length divided by the difference of the speeds. Two such conditions give two equations in the two speeds, which are solved simultaneously.