Example Question - distance formula
Here are examples of questions we've helped users solve.
Calculation of Distance Travelled by Bus
这道题是关于速度时间距离的问题。我们先把题目翻译成英文，然后求解。 题目内容是：“一辆公共汽车以每小时50公里的速度行驶，连续行驶了一小时38分钟，共行驶了多少公里？” 我们先将1小时38分钟转换成小时数，因为速度的单位是每小时（公里/小时）。 1小时 = 1小时 38分钟 = 38/60小时 所以，总时间是： 1小时 + 38/60小时 = 1 + 38/60 = 1 + 19/30 = 1 + 0.6333... ≈ 1.63小时 公式为： 距离 = 速度 × 时间 将给定的速度和时间代入公式中来计算距离： 距离 = 50公里/小时 × 1.63小时 ≈ 81.5公里 所以，公共汽车总共行驶了大约81.5公里。
Calculation of Remaining Distance and Time
Đầu tiên, chúng ta cần tính toán quãng đường mà người đó đi được sau 30 phút đầu tiên với vận tốc 36km/h. Đổi 30 phút thành giờ để tính toán dễ dàng hơn: 30 phút = 30/60 giờ = 0,5 giờ Quãng đường đi được sau 30 phút (S1) bằng vận tốc nhân với thời gian: S1 = vận tốc * thời gian = 36km/h * 0,5 giờ = 18km Bây giờ chúng ta tính quãng đường còn lại mà người đó sẽ phải đi từ điểm sau 30 phút đầu tiên đến B, với vận tốc 51km/h. Tổng quãng đường từ A đến B là 36km. Quãng đường còn lại (S2) để đến B là: S2 = Tổng quãng đường - Quãng đường đã đi = 36km - 18km = 18km Bây giờ, ta sẽ tính thời gian để đi hết quãng đường còn lại với vận tốc 51km/h. Thời gian cần thiết (t2) bằng quãng đường chia cho vận tốc: t2 = S2 / vận tốc mới = 18km / 51km/h Thực hiện phép chia: t2 = 18km / 51km/h = 0.3529 giờ (chúng ta giữ một số chữ số sau dấu phẩy để đảm bảo tính chính xác) Đổi t2 từ giờ sang phút để thể hiện kết quả một cách rõ ràng hơn: t2 = 0.3529 giờ * 60 phút/giờ ≈ 21.18 phút Vậy sau khi tăng vận tốc lên 51km/h, người đó sẽ mất khoảng 21 phút để đến B từ điểm họ đã đi được sau 30 phút đầu tiên. Rút gọn, ta có thể nói người đó sẽ cần khoảng 21 phút để hoàn thành phần đường còn lại đến B.
Calculating Distance Traveled
Đề bài viết: "Một người đi xe máy từ A đến B mất 3 giờ 30 phút. Sau đó người đó đi từ B đến A mất 4 giờ. Biết rằng tốc độ từ về chậm hơn tốc độ đi là 5 km/h. Tính quãng đường AB." Để giải bài toán này, ta áp dụng công thức tính quãng đường: Quãng đường = Tốc độ x Thời gian. Gọi quãng đường AB là S (km), tốc độ khi đi từ A đến B là v (km/h) và thời gian đi từ A đến B là 3 giờ 30 phút (tức 3,5 giờ) thì ta có phương trình: S = v x 3,5 (1) Khi quay trở lại từ B đến A, tốc độ chậm hơn 5 km/h nên tốc độ lúc này là (v - 5) km/h và thời gian mất là 4 giờ, ta có: S = (v - 5) x 4 (2) Từ hai phương trình trên, ta có hệ phương trình: v x 3,5 = (v - 5) x 4 Giải hệ phương trình này: 3,5v = 4v - 20 0,5v = 20 v = 20 / 0,5 v = 40 (km/h) là tốc độ khi người đó đi từ A đến B. Thay v vào một trong hai phương trình (1) hoặc (2) để tìm S: S = 40 x 3,5 = 140 (km) Vậy quãng đường AB dài 140 km.
Calculating Distance Between Two Points on a Coordinate Plane
The question is asking you to calculate the distance between two points on a coordinate plane, which essentially is asking you to find the length of the hypotenuse of a right triangle formed with these two points as its endpoints. The points provided are (3, 7) and (8, 5). To determine the distance between these two points, you can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Where (x1, y1) and (x2, y2) are the coordinates of the two points. Substitute the coordinates into the formula: \[ d = \sqrt{(8 - 3)^2 + (5 - 7)^2} \] \[ d = \sqrt{(5)^2 + (-2)^2} \] \[ d = \sqrt{25 + 4} \] \[ d = \sqrt{29} \] So, the distance between the two points to the nearest tenth is approximately \(\sqrt{29}\), which is about 5.4 when rounded to the nearest tenth.
Calculating Distance Between Two Points in a Coordinate Plane
To find the distance between two points, we can use the formula for the distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the coordinate plane: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] The points given are \( (-3, -4) \) and \( (-10, -9) \). Let's plug these coordinates into the formula: \[ d = \sqrt{(-10 - (-3))^2 + (-9 - (-4))^2} \] \[ d = \sqrt{(-10 + 3)^2 + (-9 + 4)^2} \] \[ d = \sqrt{(-7)^2 + (-5)^2} \] \[ d = \sqrt{49 + 25} \] \[ d = \sqrt{74} \] Now, let's find the square root of 74. \[ d \approx 8.602 \] Rounded to the nearest tenth, the distance is approximately 8.6 units.
Calculating Distance Between Two Points in a Plane
To find the distance between two points in a plane, you can use the distance formula, which is derived from the Pythagorean theorem. The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] For the points \( (1, -7) \) and \( (-6, 2) \), we can calculate the distance as follows: Let \( x_1 = 1 \), \( y_1 = -7 \), \( x_2 = -6 \), and \( y_2 = 2 \). \[ d = \sqrt{(-6 - 1)^2 + (2 - (-7))^2} \] \[ d = \sqrt{(-7)^2 + (9)^2} \] \[ d = \sqrt{49 + 81} \] \[ d = \sqrt{130} \] \[ d \approx 11.4 \] (rounded to the nearest tenth) So the distance between the points \( (1, -7) \) and \( (-6, 2) \) is approximately 11.4 units.
Calculating Distance Between Two Points in a Coordinate System
To find the distance between two points in a coordinate system, you can use the distance formula: Distance = √[(x2 - x1)² + (y2 - y1)²] Plugging in the coordinates for the two points (9,2) and (2,9), we get: x1 = 9, y1 = 2 x2 = 2, y2 = 9 Distance = √[(2 - 9)² + (9 - 2)²] Distance = √[(-7)² + (7)²] Distance = √[49 + 49] Distance = √[98] Now, we can round the result to the nearest tenth: Distance ≈ √[98] ≈ 9.899494937 Rounded to the nearest tenth, the distance is approximately 9.9 units.
Calculating Distance Between Two Points
To find the distance between the points (8, 2) and (3, 8), we use the distance formula derived from the Pythagorean theorem: Distance = √[(x2 - x1)² + (y2 - y1)²] Here, (x1, y1) is the point (8, 2) and (x2, y2) is the point (3, 8). Plugging in these values: Distance = √[(3 - 8)² + (8 - 2)²] Distance = √[(-5)² + (6)²] Distance = √[25 + 36] Distance = √61 Distance ≈ 7.8 (rounded to the nearest tenth) So, the distance between the points is approximately 7.8 units.
Finding the Distance between Two Points
To find the distance between the two points (9, 1) and (3, 10), you can use the distance formula, which is derived from the Pythagorean theorem: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Here, \( (x_1, y_1) = (9, 1) \) and \( (x_2, y_2) = (3, 10) \). Plugging in the values: \[ \text{Distance} = \sqrt{(3 - 9)^2 + (10 - 1)^2} \] \[ \text{Distance} = \sqrt{(-6)^2 + (9)^2} \] \[ \text{Distance} = \sqrt{36 + 81} \] \[ \text{Distance} = \sqrt{117} \] Now, simplify this to get the final answer: \[ \text{Distance} = \sqrt{117} \approx 10.82 \] The question asks to round decimals to the nearest tenth: \[ \text{Distance} \approx 10.8 \text{ units} \]
Calculating Distance Between Two Points on a Coordinate Plane
To find the distance between two points on a coordinate plane, you can use the distance formula, which is derived from the Pythagorean theorem. The distance formula is: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) The two points you have are (2, -7) and (8, -7). Plugging these into the formula gives: \( x_1 = 2, y_1 = -7 \) \( x_2 = 8, y_2 = -7 \) Now compute the distance: \( d = \sqrt{(8 - 2)^2 + (-7 - (-7))^2} \) \( d = \sqrt{(6)^2 + (0)^2} \) \( d = \sqrt{36 + 0} \) \( d = \sqrt{36} \) \( d = 6 \) So the distance between the points (2, -7) and (8, -7) is 6 units.
Solving for Initial Velocity and Distance with an Accelerated Motion Formula
The given formula is: \[ s = ut + \frac{1}{2}at^2 \] To solve part (a), we want to make \( u \) the subject of the formula. Here's how we do it: First, let's rearrange the formula to solve for \( u \) by subtracting \( \frac{1}{2}at^2 \) from both sides of the equation: \[ s - \frac{1}{2}at^2 = ut \] Next, we divide both sides of the equation by \( t \) to solve for \( u \): \[ \frac{s - \frac{1}{2}at^2}{t} = u \] So the formula with \( u \) as the subject is: \[ u = \frac{s - \frac{1}{2}at^2}{t} \] For part (b), we need to calculate \( s \) when \( a = 100 \text{ m/s}^2 \), \( u = 2 \text{ m/s} \), and \( t = 5 \text{ s} \). We can plug the given values into the original formula: \[ s = ut + \frac{1}{2}at^2 \] \[ s = (2 \text{ m/s})(5 \text{ s}) + \frac{1}{2}(100 \text{ m/s}^2)(5 \text{ s})^2 \] \[ s = 10 \text{ m} + \frac{1}{2}(100 \text{ m/s}^2)(25 \text{ s}^2) \] \[ s = 10 \text{ m} + 50(25 \text{ m}) \] \[ s = 10 \text{ m} + 1250 \text{ m} \] \[ s = 1260 \text{ m} \] Thus, the distance \( s \) is 1260 meters.
Analyzing a Quadrilateral Using Coordinates and Slopes
It seems that you're requesting help with a math problem involving the coordinates of a quadrilateral. Unfortunately, the image is not clear enough for me to provide a definitive answer, but I can still guide you on how to solve this type of problem. We have a quadrilateral ABCD with points A(2,4), B(4,2), C(-1,-3), and D(3,-1). To answer the questions, you need to: a. Find the slope of each side of quadrilateral ABCD: - The slope of a line through points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1). For AB: slope = (2 - 4) / (4 - 2) = -2 / 2 = -1 For BC: slope = (-3 - 2) / (-1 - 4) = -5 / -5 = 1 For CD: slope = (-1 - (-3)) / (3 - (-1)) = 2 / 4 = 0.5 For DA: slope = (4 - (-1)) / (2 - 3) = 5 / -1 = -5 b. What type of quadrilateral is formed by the fences? Justify your answer using the slope and/or distance formulas: - To determine the type of quadrilateral, we need to check for parallel sides (equal slopes) and equal lengths (using the distance formula). The distance formula for points (x1, y1) and (x2, y2) is given by √[(x2 - x1)^2 + (y2 - y1)^2]. Find the distance of each side: AB = √[(4 - 2)^2 + (2 - 4)^2] = √[2^2 + (-2)^2] = √(4 + 4) = √8 BC = √[(-1 - 4)^2 + (-3 - 2)^2] = √[(-5)^2 + (-5)^2] = √(25 + 25) = √50 CD = √[(3 - (-1))^2 + (-1 - (-3))^2] = √[4^2 + 2^2] = √(16 + 4) = √20 DA = √[(2 - 3)^2 + (4 - (-1))^2] = √[(-1)^2 + 5^2] = √(1 + 25) = √26 The sides AB and CD are not parallel (their slopes are not equal), and sides BC and DA are also not parallel. Also, none of the side lengths are equal. Without plotting the points, we can't be entirely certain what specific type of quadrilateral it is, but we can confirm that it is not a parallelogram, rectangle, square, or rhombus since none of the sides are equal in length or parallel. It's most likely a general quadrilateral with no special properties regarding its sides or angles. Keep in mind that the calculations for the distances of sides AB, BC, CD, and DA should be recalculated with proper care for precision and to ensure accuracy in the answer. If the points were plotted on a graph, the shape could also be visually identified and confirmed.
Determining Distance to Dog Park
The image provides a word problem about James and Amber who walk their dogs together at a nearby dog park. They wish to determine who has to walk a farther distance to get to the dog park. The question requires the use of a coordinate plane to determine the distances, with the town square (origin) at (0,0), James's house at (1, -4), Amber's house at (2, -6), and the dog park at (2, 4). To find out how far each person lives from the dog park, you need to calculate the distance between their houses and the dog park using the distance formula for coordinates: Distance = √((x2 - x1)² + (y2 - y1)²) For James: Distance between James's house and the dog park = √((2 - 1)² + (4 - (-4))²) = √((1)² + (8)²) = √(1 + 64) = √(65) For Amber: Distance between Amber's house and the dog park = √((2 - 2)² + (4 - (-6))²) = √((0)² + (10)²) = √(0 + 100) = √(100) = 10 Comparing the two distances, √(65) is less than 10, which means James lives closer to the dog park than Amber. The statement to complete would therefore be: James is √(65) blocks from the dog park and Amber is 10 blocks from the dog park. This means that Amber is farther from the dog park.
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