If {name} drives at an average rate of @a {speed}, how long will it them to drive @b {dim}?
The core equation here is that {distance=rate*time} Let d=distance, r=rate, and t=time, then {d=rt} Here you are given the rate, {r=@a {speed}} and the distance {d=@b {dim}} and you want to find t. Solve Equation 2 for t to get {t=d/r} Plug in your numbers to find that {t=@b/@a}
If {name} drives at an average rate of @a {speed} for @b {time}. How far will they travel?
The core equation here is that {distance=rate*time} Let d=distance, r=rate, and t=time, then {d=rt} Here you are given the rate, {r=@a {speed}} and the time they traveled at this rate {t=@b {time}} and you want to find d, the distance traveled. Plug these numbers into {d=rt} to find that {d=(@a)(@b)}
If {name} drives @a {dim} in @b {time}. How fast did they travel?
The core equation here is that {distance=rate*time} Let d=distance, r=rate, and t=time, then {d=rt} Solve this for r (the rate) to get {r=d/t} Here you are given the distance, {d=@a {dim}} and the time they traveled at this rate {t=@b {time}} and you want to find r, the rate of travel. Plug these numbers into {r=d/t} to find that {r=(@a)/(@b)}
{name} drives to the next city in @a {time} in heavy traffic. When there is no traffic the same trip takes @b {time}. It is possible to drive @c {speed} faster when there is no traffic than when there is heavy traffic. What is their average speed when there is no traffic? What is the distance to the next city?
Let r=the rate when there in heavy traffic. Then, r+@c=the rate when there is no traffic. We know {distance=rate*time} or {d=rt} Let d=the distance to the nearest city. In heavy traffic, using this relationship, {d=(r)(@a)} When there is no traffic {d=(r+10)@b} Now you have two equations and two unknowns. Plug the 'd=' in Equation 3 in for d in Equation 4 and solve for r. This will be the rate in heavy traffic. Add @c to this rate (r+@c) to get the rate when there is no traffic. Plug the r into Equation 3 and compute d. This is the distance to the next city.
Person 1 leaves a city at @a {clock1} heading for another city. Person 2 leaves at @b {clock2} on the same highway, heading towards the same city. By driving @c {speed} faster, Person 2 overtakes Person 1 at @d {clock3}. Find Person 1's rate. Find Person 2's rate. Find the distance each traveled before they met.
The core equation we'll use here is that distace=rate*time or {d=rt} Let x=Person 1's distance at any given time, and y=Person 2's distance at any given time. Let r=Person 1's rate, then r+@c=Person 2's rate. Formulating d=rt on each person, we get that {x=rt} and {y=(r+@c)t} When Person 1 overtakes Person 2, Person 1 has driven @d {clock3}-@a {clock1} hours, and Person 2 has driven @d {clock3}-@b {clock1} hours. their distances must be the same, or x=y, in which case we get {(r)(@d {clock3} -@a {clock1})=(r+@c)(@d {clock3}-@b {clock2})} Solve this equation for r to get Person 1's rate. r+@c will be Person 2's rate. With r, you can use either Equation 1 or 2 to find the distance they traveled when they met. If you use Equation 1, use the time {t=@d {clock3} - @a {clock1}} If you use Equation 2, use the time {t=@d {clock3} - @b {clock2}}<p> Caution: Be careful how you subtract times of the day here. For example, between 11 AM - 7 AM, 4 hours have gone by. But, for example, 2 PM - 11AM IS NOT (-9, or 9, etc.), rather it's 3 hours that have gone by. It's the 'hours that have gone by' which you should use in the equations above.
A boat cruises downstream for @a {time} before heading back. It takes @b {time} going upstream to get back. If the speed of the stream is @c {speed}, find the speed of the boat in still water.
The core equation to use here is distance=rate*time or {d=rt} Let d=the distance traveled by the boat, which is the same both upstream and downstream (since it ends up right where it started at the end of the cruise). Let r=the speed of the boat in still water. Let s=the speed of the stream, so s=@c {speed}. Downstream, the boat travels with the water, hence faster, so its rate is {r+s} Upstream, the boat travels slower, against the water, so its rate is slower or {r-s} So the distance downstream that the boat travels is {d=(r+@c)(@a)} and the distance upstream that the boat travels is {d=(r-@c)(@b)} But these two distance are the same (again, since the boat ends up where it started), so we can set Equation 4 and 5 equal to each other, or {(r+@c)(@a)=(r-@c)(@b)} Solve for r to get the speed of the boat in still water.
Two planes are @a {dim} apart. They fly towards each other, one at @b {speed} and the other at @c {speed}. How long does it take the planes to pass each other?
The key equation here is distance=rate*time, or d=rt. Let's reference all distances from the point where the plane traveling at @b {speed} takes off. Its distance from its take-off point at any time is {d=@bt} The distance of the plane traveling at @c {speed} from this same take-off point at any time is {d=@a-@ct} When they pass each other, these two 'd=' equations are equal to each other (meaning the planes have the same position or distance from our take-off reference point). So {@bt=@a-@ct} Solve for t to find the time at which the planes will pass each other.
Two cars start at the same time from cities that are @a {dim} apart and travel towards each other. The rate of the faster car exceeds the rate of the slower car by @b {speed}. After @c hours the cars were still @d {dim} apart. Find the rate of each car.
The key equation here is distance=rate*time, or d=rt. Let r=the rate of the slower car. The rate of the faster car will be r+@b. Let's reference all distances from the city where the slower car starts off. At any time, its distance from the city is {d=rt} The distance the faster car is from this same city at any given time is {d=360-(r+@b)t} After @c hours, the distance between the cars is @d {dim}. The distance between the cars is just the difference between their two distance equations, given in Equations 1 and 2. Subtracting these, we get {360-(r+@b)t-rt} When t=@c, this difference is to be @d {dim}, so {360-@c(r+@b)-@cr=@d} Solve this for r to get the rate of the slower car. Add @b to this answer to get the rate of the faster car.
{name} rides @a hours on a bicycle trip into the country and back. They rode out at the rate of @b {speed} and returned at the rate of @c {speed} How far into the country did they ride?
We'll use the fact that distance=rate*time here, or d=rt. Solving this for t, we get that {t=d/r} The time it took {name} to ride out is {t=d/@b} The time it took {name} to ride back is {t=d/@c} The total time of the trip is stated at @a hours. But the total time is given by the time it takes to ride out plus the time it takes to ride back, which is the t in Equation 2 plus the t in Equation 3. Adding these, we get that {d/@b+d/@c=@a} Solve this for d to find out how far {name} rode out into the country.
At @a {clock1} a freight train leaves town traveling at @b {speed}. At @c {clock2} (a later time) a passenger train leaves the same station traveling in the same direction, along the same route at @d {speed}. How long will it take the passenger train to overtake the freight train? How far will they be from the station at this time?
We'll use the fact that distance=rate*time here, or d=rt. The first thing we need to do is figure out the difference in the leaving times between the passenger and freight trains. This problem assumes you chose times that mean the passenger train leave after the freight train. So whatever this difference is (in hours) it will be something like {@c {clock2} - @a {clock1}} Figure out how many hours are between these two times of day, and call the result h, so {h=@c {clock2} - @a {clock1}} in hours. Now the distance the freight train is from the station at any time after the passenger train leaves is {d=@b(t+h)} Notice the t+h term in this equation, indicating that the freight train had a head-start (h for head-start). The distance the passenger train is from the station is {d=@dt} When they pass each other the two trains' distances from the station is the same, or {@b(t+h)=@dt} Solve this for t to find how long it will take the trains to pass each other. Plug this t into Equation 3 or 4 to find the distance from the station this overtaking occurs.
Two trains started at the same time from the same station and traveled in opposite directions. One traveled at a rate of @a {speed} and the other at the rate of @b {speed}. In how many hours were they @c {dim} apart?
We'll use the fact that distance=rate*time here, or d=rt. The distance one train is from the station after some time t is {d=@at} The distance the other train is from the station after the same time t is {d=@bt} At any given time, the distance between the two trains can be found by adding the d's in Equations 1 and 2 together like this {@at+@bt} This is the distance between the trains at any given time. When they are @c {dim} apart, {@at+@bt=@c} Solve this for t to find the number of hours it takes before the trains are @c {dim} apart.
