Lesson Entertainment Travel and Distance problems

Entertainment Travel and Distance problems

Problem 1

Tom's   rate is  = 240 m/min.

Jerry's rate is  = 200 m/min


Tom is running faster, so he will make his 6000 m and will be in his returning way when he will meet Jerry.

So, the equations are 

6000 + x = 240t   (1)   for Tom, and
6000 - x = 200t   (2)   for Jerry.

"x" is "additional distance for Tom, t is the time before they meet each other.

Add equations (1) and (2). You will get

12000 = 240t + 200t,  or 12000 = 440t,  hence,  t =  = 27.(27) minutes before they meet each other.


Answer.  The time they've each been running for when they meet each other is  27.(27) minutes.

Problem 2

The problem says that Ryan, who is now 60 feet behind Nelson, should catch him up in 30 seconds.


So, Ryan should running in  ft/s = 2 ft/s faster than Nelson, who runs at 6 ft/s.


Hence, Ryan' speed must be 6 + 2 = 8 ft/s.


Answer.  Ryan' speed must be 6 + 2 = 8 ft/s.

Problem 3

As everybody knows, the hour hand makes one full rotation in 12 hours.

Hence, the regular angular speed of the hour hand is  of 360 degs, or 30 degs per hour, or  0.5 degs per minute.


Again, as everybody knows, the minute hand makes one full rotation in 1 hour = 60 minutes. 

Hence, the regular angular speed of the minute hand is 360 degs per hour,  or 6 degs per minute.


The equation for the angular position of the "crazy" hour hand is

    H(t) = 90 - (6*0.5)*t                     (1)

where 90 = 90 degs is its initial position in degrees at noon ("vertically up");  (6*0.5) is its "crazy" angular velocity 
and "t" is time (! the CORRECT time !).


The equation for the angular position of the "crazy" minute hand is

    M(t) =        (2)

where 90 = 90 degs is its initial position in degrees at noon ("vertically up");  ((5/6)*6) is its "crazy" angular velocity and 
"t" is time in minutes (! the CORRECT time !). ( ! Notice that the sign "+" in formula (2) reflects anti-clockwise rotation). 


The condition that the hour and the minute hands coincide is

    M(t) = H(t) + 360 degs,   or    = 90 - (6*0.5)*t + 360   (3)


Simplify equation (3) step by step.

    5*t = -3*t + 360  ====>  5t + 3t = 360  ====>  8t = 360  ====>  t =  = 45 minutes.

       The method of solving the problem for "crazy" clock is the same as for normal/regular clock.

       See the lesson  Advanced clock problems.

Problem 4

Joe and Frank approach each other at the rate 6+6 = 12 miles per hour.


Hence, they will meet each other in 36/12 = 3 hours.


During these 3 hours, flying at the rate of 18 miles per hour, the pigeon will cover the distance of 3*18 = 54 miles.   ANSWER

Problem 5

Let x be the Susan's rate (in meters per second), and let y be the Jenny's rate.

Susan's time is    seconds;  it is equal to  , according to the condition.


So, we have   = ,  or    =  = .   (1)


Next, in the other scenario, Susan's time to complete 210 m run will be    seconds,
while the Jenny's time to complete her run of 200 m  will be  .


The problem asks which value is lesser,    or  .


From (1),  we have x = ,  therefore

    Susan's time will be   =  =  =  = .

    Obviously, it is less than the Jenny's time  .


So, the conclusion is:  Susan will win (will be first) in the second scenario run.    ANSWER

There is even more easy and comprehensive explanation.