Rate, Distance, and Time Problems, 8

Category: Physics, Mechanics

"Published in Newark, California, USA"

A runner A can run the mile race in 4.25 min. Another runner B requires 4.55 min to run this distance. If they start out together and maintain their normal speeds, how far apart will they be at the finish of the race?

Solution:

The given problem is about rate, distance, and time problem in which the two runners run in the same mile race. From the given problem also, runner A is faster than runner B because runner A has a lesser time to complete the mile race than runner B.

The speed of runner A is

The speed of runner B is

If the two runners will start together in a mile race, then runner A will finish the race. Therefore, their distance between the two runners at the end of the race is

For runner B, we use the time for runner A because runner A will finish the race.

                        or

 

Rate, Distance, and Time Problems, 7

Category: Physics, Mechanics

"Published in Newark, California, USA"

A motorist has to travel 3.50 km in a city where his average speed should not exceed 25 km/hr. If he increases his average speed to 40 km/hr, how much time will he gain in his journey?

Solution:

The given problem is about rate, distance, and time problem in which the average speed is increased. Let's assume that the distance for the two cases is the same which is 3.50 km. 

If the speed limit in a city is 25 km/hr, then his travel time is

If he increased his speed to 40 km/hr, then his travel time is
 

 

 

 

Therefore, his travel time will gain by
 

 

 

                         or

 

Index of Refraction Problems, 5

Category: Physics

"Published in Newark, California, USA"

 A coin is placed beneath a rectangular slab of glass 5.0 m thick. If the index of refraction of the glass is 1.500, how far beneath the upper surface of the glass will the coin appear to be when viewed vertically from above?

Solution:

The given problem is about the refraction of light in which the incident ray and refracted ray are coincide with the normal line. To illustrate the problem, it is better to draw the figure as follows  

Photo by Math Principles in Everyday Life

Since the index of refraction of a glass is already given in the problem which is n = 1.500, therefore, the apparent depth of a coin from the upper surface of a glass is