Rate, Time, and Distance Problems – Solved Examples with Step-by-Step Solutions

After \(t\) hours, the distances \(D_1\) and \(D_2\), in miles per hour, travelled by the two cars are given by \[ D_1 = 40t \quad \text{and} \quad D_2 = 50t \] After \(t\) hours, the distance \(D\) separating the two cars is given by \[ D = D_1 + D_2 = 40t + 50t = 90t \] The distance \(D\) will be equal to 450 miles when \[ D = 90t = 450 \] To find the time \(t\) for \(D\) to be 450 miles, solve the above equation for \(t\) to obtain \[ t = 5 \text{ hours.} \] Therefore, if the cars start at 5 am, they will be 450 miles apart at \[ 5 \text{ am} + 5 \text{ hours} = 10 \text{ am.} \]

After \(t\) hours, the distance \(D_1\) traveled by car A is given by \[ D_1 = 40 t \] Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it. After \((t - 1)\) hours, the distance \(D_2\) traveled by car B is given by \[ D_2 = 60 (t - 1) \] When car B passes car A, they are at the same distance from the starting point, and therefore \(D_1 = D_2\), which gives \[ 40 t = 60 (t - 1) \] Solving the above equation for \(t\) gives \[ t = 3 \text{ hours} \] Car B passes car A at \[ 9 + 3 = 12 \text{ pm} \]

After \( t \) hours, the two trains will have traveled distances \( D_1 \) and \( D_2 \) (in miles) given by \[ D_1 = 72 t \quad \text{and} \quad D_2 = 78 t \] After \( t \) hours, the total distance \( D \) traveled by the two trains is given by \[ D = D_1 + D_2 = 72 t + 78 t = 150 t \] When the distance \( D \) is equal to 900 miles, the two trains pass each other: \[ 150 t = 900 \] Solve the above equation for \( t \): \[ t = 6 \text{ hours.} \]

The total distance \( D \) traveled by John is given by \[ D = 45 \times 2 + 3 \times 55 = 255 \text{ miles}. \]

If \(x\) is the rate at which Linda drove before lunch, the rate after lunch is equal to \(x + 10\). The total distance \(D\) traveled by Linda is given by \[ D = 2x + 3(x + 10) \] and is equal to 230 miles. Hence \[ 2x + 3(x + 10) = 230 \] Solving for \(x\) gives \[ x = 40 \text{ miles/hour}. \]

The two cars are traveling in directions that are at right angles. Let \(x\) and \(y\) be the distances traveled by the two cars in \(t\) hours. Hence: \[ x = 50 t \quad \text{and} \quad y = 60 t \] Since the two directions are at right angles, Pythagoras' theorem can be used to find the distance \(D\) between the two cars as follows: \[ D = \sqrt{x^2 + y^2} \] We now find the time at which \(D = 300\) miles by solving: \[ \sqrt{x^2 + y^2} = 300 \] Square both sides and substitute \(x\) and \(y\) by \(50 t\) and \(60 t\) respectively to obtain the equation: \[ (50 t)^2 + (60 t)^2 = 300^2 \] Solve the above equation to obtain: \[ t = 3.84 \text{ hours (rounded to two decimal places) or 3 hours and 51 minutes (to the nearest minute)} \] The two cars will be 300 miles apart at: \[ 8 + 3 \text{ h } 51' = 11:51 \text{ am} \]

Let \(x\) be John's rate in traveling between the two cities. The rate of Peter will be \(x + 20\). We use the rate-time-distance formula to write the distance \(D\) traveled by John and Peter (same distance \(D\)): \[ D = 3x \quad \text{and} \quad D = 2(x + 20) \] The first equation can be solved for \(x\) to give: \[ x = \frac{D}{3} \] Substitute \(x\) by \(\frac{D}{3}\) into the second equation: \[ D = 2\left(\frac{D}{3} + 20\right) \] Solve for \(D\) to obtain: \[ D = 120 \text{ miles} \]

Let \(D\) be the distance between the two cities. When Gary and Thomas cross each other, they have covered all the distance between the two cities. Hence \[ D_1 = 2 \times 50 = 100 \text{ miles}, \quad \text{distance traveled by Gary} \] \[ D_2 = 2 \times (50 + 15) = 130 \text{ miles}, \quad \text{distance traveled by Thomas} \] The distance \(D\) between the two cities is given by \[ D = D_1 + D_2 = 100 \text{ miles} + 130 \text{ miles} = 230 \text{ miles}. \]

Let \(D_1\) and \(D_2\) be the distances traveled by the two cars in \(t\) hours. \[ D_1 = 50t \quad \text{and} \quad D_2 = 60t \] The second car has a higher speed, so the distance \(d\) between the two cars is given by \[ d = 60t - 50t = 10t \] For \(d\) to be 30 miles, we have \[ 30 = 10t \] Solving the above equation for \(t\) gives \[ t = 3 \text{ hours} \]

Let \(D_1\) and \(D_2\) be the distances traveled by the two trains in \(t\) hours. \[ D_1 = 80 t \quad \text{and} \quad D_2 = 100 t \] Since the two trains are traveling in opposite directions, the total distance \(D\) between the two trains is given by \[ D = D_1 + D_2 = 180 t \] For this distance to be 450 miles, we have \[ 180 t = 450 \] Solving for \(t\) gives \[ t = 2.5 \text{ hours} = 2 \text{ hours } 30 \text{ minutes} \] Thus, if the first train leaves at 10:00 pm, the trains will be 450 miles apart at \[ 10:00 \text{ pm } + 2:30 = 12:30 \text{ am}. \]

When the first train has traveled for \(t\) hours, the second train will have traveled \((t - 1)\) hours since it started 1 hour late. Hence, if \(D_1\) and \(D_2\) are the distances traveled by the two trains, then \[ D_1 = 80 t \quad \text{and} \quad D_2 = 100 (t - 1) \] Since the trains are traveling in opposite directions, the total distance \(D\) between the two trains is given by \[ D = D_1 + D_2 = 180 t - 100 \] For \(D\) to be 530 miles, we need to have \[ 180 t - 100 = 530 \] Solving for \(t\) gives \[ t = 3 \text{ hours } 30 \text{ minutes} \] Adding this to the start time: \[ 8:00 \text{ am } + 3:30 = 11:30 \text{ am} \]