Two cars start from a point at the same time and travel on two different roads at right angles to each other. Their speed is 18 km/h and 72 km/h respectively. What will be the distance between them after 6 seconds?
Understanding the Problem: Finding Distance Between Cars at Right Angles
This problem involves two cars starting from the same point and moving in directions perpendicular to each other. This means their paths form the two perpendicular sides of a right-angled triangle. The distance between the cars after a certain time will be the hypotenuse of this triangle.
We are given the speeds of the cars in kilometers per hour (km/h) and the time in seconds. To calculate the distances traveled, we first need to convert the speeds to meters per second (m/s).
The conversion factor from km/h to m/s is \( \frac{5}{18} \).
Step 1: Convert Speeds from km/h to m/s
- Speed of Car 1 = 18 km/h
- Speed of Car 2 = 72 km/h
Converting Speed of Car 1:
\( \text{Speed}_1 = 18 \, \text{km/h} \times \frac{5}{18} \, \text{m/s per km/h} = 5 \, \text{m/s} \)
Converting Speed of Car 2:
\( \text{Speed}_2 = 72 \, \text{km/h} \times \frac{5}{18} \, \text{m/s per km/h} = 4 \times 5 \, \text{m/s} = 20 \, \text{m/s} \)
Step 2: Calculate Distance Traveled by Each Car
The cars travel for 6 seconds. We use the formula: Distance = Speed × Time.
- Time = 6 seconds
Distance traveled by Car 1 (\(d_1\)):
\( d_1 = \text{Speed}_1 \times \text{Time} = 5 \, \text{m/s} \times 6 \, \text{s} = 30 \, \text{meters} \)
Distance traveled by Car 2 (\(d_2\)):
\( d_2 = \text{Speed}_2 \times \text{Time} = 20 \, \text{m/s} \times 6 \, \text{s} = 120 \, \text{meters} \)
Step 3: Calculate the Distance Between the Cars Using the Pythagorean Theorem
Since the roads are at right angles, the distances \(d_1\) and \(d_2\) form the perpendicular sides of a right-angled triangle. The distance between the cars (\(D\)) is the hypotenuse.
According to the Pythagorean theorem:
\( D^2 = d_1^2 + d_2^2 \)
Substitute the calculated distances:
\( D^2 = (30)^2 + (120)^2 \)
\( D^2 = 900 + 14400 \)
\( D^2 = 15300 \)
Now, take the square root to find \(D\):
\( D = \sqrt{15300} \)
To simplify the square root, we can factorize 15300:
\( 15300 = 100 \times 153 \)
\( 153 = 9 \times 17 \)
So, \( 15300 = 100 \times 9 \times 17 \)
\( D = \sqrt{100 \times 9 \times 17} = \sqrt{100} \times \sqrt{9} \times \sqrt{17} \)
\( D = 10 \times 3 \times \sqrt{17} \)
\( D = 30\sqrt{17} \) meters
Therefore, the distance between the two cars after 6 seconds will be \(30\sqrt{17}\) meters.
Summary of Distances and Calculations
| Item | Value | Calculation/Notes |
|---|---|---|
| Speed Car 1 (km/h) | 18 | Given |
| Speed Car 2 (km/h) | 72 | Given |
| Time (s) | 6 | Given |
| Speed Car 1 (m/s) | 5 | \(18 \times \frac{5}{18}\) |
| Speed Car 2 (m/s) | 20 | \(72 \times \frac{5}{18}\) |
| Distance Car 1 (m) | 30 | \(5 \times 6\) |
| Distance Car 2 (m) | 120 | \(20 \times 6\) |
| Distance Between Cars (m) | \(30\sqrt{17}\) | \(\sqrt{30^2 + 120^2} = \sqrt{900 + 14400} = \sqrt{15300}\) |
The calculated distance matches one of the provided options.
Important Questions from Lines and Angles
Two parallel lines are intersected by a transversal, the corresponding angles are:
If l, m, n are lines such that, l is parallel to n and m is parallel to n, then, ______
There are three points P, Q and R on a straight line such that PQ : QR = 3 : 5. If n is the number of possible values of PQ : PR, then what is n equal to ?
If angles of a triangle are in the ration of 2 : 3 : 4, then the measure of the smallest angle is:
If the angles of a triangle are in the ratio of 2 : 3 : 5, then find the ratio of the greatest angle to the smallest angle.
A. 7 : 2
B. 5 : 2
C. 5 : 3
D. 3 : 5
