**Motion Problems**

Virtually, all motion problems involve the formula Distance = Rate Ã— Time, or D = R Ã— T

**Overtake**:

In this type of problem, one person catches up with or overtakes another person. The key to these problems is that at the moment one person overtakes the other they have traveled the same distance.

**Example:**

Scott starts jogging from point X to point Y. A half-hour later his friend Garrett who jogs 1 mile per hour slower than twice ScottÂ’â€™s rate starts from the same point and follows the same path. If Garrett overtakes Scott in 2 hours, how many miles will Garrett have covered?

(A) $2\frac{1}{5}$

(B) $3\frac{1}{3}$

(C) 4

(D) 6

(E) $6\frac{2}{3}$

Following Guideline 1, we let r = *Scott's rate.* Then 2r Â–â€“ 1 = *Garrett's rate.* Turning to Guideline 2, we look for two quantities that are equal to each other. When Garrett overtakes Scott, they will have traveled the same distance. Now, from the formula D = R Ã— T , ScottÂ’â€™s distance is D = r Ã— 2 $\frac{1}{2}$ and GarrettÂ’â€™s distance is D = (2r Â–- 1)2 = 4r Â–- 2

Setting these expressions equal to each other gives 4r - 2 = r Ã— 2 $\frac{1}{2}$

Solving this equation for r gives us r = 4/3

Hence, Garrett will have traveled D = 4r - 2 = 4( $\frac{4}{3}$ ) -2 = 3 $\frac{1}{3}$ miles. The answer is (B).

**Opposite Directions:**In this type of problem, two people start at the same point and travel in opposite directions. The key to these problems is that the total distance traveled is the sum of the individual distances traveled.

**Example:**

Two people start jogging at the same point and time but in opposite directions. If the rate of one jogger is 2 mph faster than the other and after 3 hours they are 30 miles apart, what is the rate of the faster jogger?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

Let r be the rate of the slower jogger. Then the rate of the faster jogger is r + 2. Since they are jogging for 3 hours, the distance traveled by the slower jogger is D = rt = 3r, and the distance traveled by the faster jogger is 3(r + 2). Since they are 30 miles apart, adding the distances traveled gives

3r + 3(r + 2) = 30

3r + 3r + 6 = 30

6r + 6 = 30

6r = 24

r = 4

Hence, the rate of the faster jogger is r + 2 = 4 + 2 = 6. The answer is (D).

**Round Trip:**The key to these problems is that the distance going is the same as the distance returning.

**Example: **

A cyclist travels 20 miles at a speed of 15 miles per hour. If he returns along the same path and the entire trip takes 2 hours, at what speed did he return?

(A) 15 mph

(B) 20 mph

(C) 22 mph

(D) 30 mph

(E) 34 mph

Solving the formula D= R Ã— T for T yields T =$\frac{D}{R}$. For the first half of the trip, this yields T = $\frac{20}{15}$ = $\frac{4}{3}$ hours. Since the entire trip takes 2 hours, the return trip takes 2 - $\frac{4}{3}$ hours, or $\frac{2}{3}$ hours. Now, the return trip is also 20 miles, so solving the formula D= R Ã— T for R yields R = $\frac{D}{T}$ = $\frac{20}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$ = 20 Â· $\frac{3}{2}$ = 30.

The answer is (D).

**Compass Headings:**In this type of problem, typically two people are traveling in perpendicular directions. The key to these problems is often the Pythagorean Theorem.

**Example: **

At 1 PM, Ship A leaves port heading due west at x miles per hour. Two hours later, Ship B is 100 miles due south of the same port and heading due north at y miles per hour. At 5 PM, how far apart are the ships?

(A) âˆš(4 x)^{2} + (100 + 2y)^{2}

(B) x + y

(C) âˆšx^{2} + y2

(D) âˆš(4 x)^{2} + (2y)^{2}

(E)âˆš(4 x)^{2} + (100 - 2y)^{2}

Since Ship A is traveling at x miles per hour, its distance traveled at 5 PM is D = rt = 4x. The distance traveled by Ship B is D = rt = 2y. This can be represented by the following diagram:

Applying the Pythagorean Theorem yields s ^{2} = (4x)^{2} + (100 - 2y)^{2} . Taking the square root of this equation gives s = âˆš(4 x)^{2} + (100 - 2y)^{2} .

The answer is (E).

**Circular Motion:**In this type of problem, the key is often the arc length formula S = RÎ¸, where S is the arc length (or distance traveled), R is the radius of the circle, and Î¸ is the angle.

**Example 1: **

The figure to the right shows the path of a car moving around a circular racetrack. How many miles does the car travel in going from point A to point B ?

(A) Ï€/6

(B) Ï€/3

(C) Ï€

(D) 30

(E) 60

When calculating distance, degree measure must be converted to radian measure. To convert degree measure to radian measure, multiply by the conversion factor $\frac{\mathrm{\xcf\u20ac}}{180}$ . Multiplying 60Â° by $\frac{\mathrm{\xcf\u20ac}}{180}$ yields 60 Â· $\frac{\mathrm{\xcf\u20ac}}{180}$ = $\frac{\mathrm{\xcf\u20ac}}{3}$ . Now, the length of arc traveled by the car in moving from point A to point B is S. Plugging this information into the formula S = RÎ¸ yields S = $\frac{1}{2}$ Â· $\frac{\mathrm{\xcf\u20ac}}{3}$ = $\frac{\mathrm{\xcf\u20ac}}{6}$Â·The answer is (A).

**Example 2: **

If a wheel is spinning at 1200 revolutions per minute, how many revolutions will it make in t seconds?

(A) 2t

(B) 10t

(C) 20t

(D) 48t

(E) 72t

Since the question asks for the number of revolutions in *t* seconds, we need to find the number of revolutions per second and multiply that number by* t*. Since the wheel is spinning at 1200 revolutions per minute and there are 60 seconds in a minute, we get $\frac{\mathrm{1200\; revolution}}{\mathrm{60\; seconds}}$ = 20 rev/sec . Hence, in *t* seconds, the wheel will make 20*t* revolutions.

The answer is (C).