Explanation

In this “chase” problem, the two vehicles are moving in the same direction, with one chasing the other. To determine how long it will take the rear vehicle to catch up, subtract the rates to find out how quickly the rear vehicle is gaining on the one in front.

The police car gains on the train at a rate of 80 – 50 = 30 miles per hour. Since the police car needs to close a gap of 50 miles, plug into the D = RT formula to find the time:

50 = 30t
\(\frac{5}{3} = t\)

The time it takes to catch up is \(\frac{5}{3}\) hours, or 1 hour and 40 minutes.

When elements compete, SUBTRACT THEIR RATES.

The rate for the police = 80 mph, while the rate for the gang = 50 mph.
Thus, every hour the police travel 80 miles, while the gang travels only 50 miles.
The difference between their rates = 80-50 = 30 mph.
Implication of this rate difference:
Every hour the police travel 30 more miles than the gang.
As a result, every hour the police CATCH-UP by 30 miles.

Since the police must catch up by 50 miles, and their catch-up rate is 30 mph, we get:
\(Catch-up-time = \frac{catch-up-distance}{catch-up-rate} = \frac{50}{30} = \frac{5}{3}\) hours = 1 hour and 40 minutes.



Alternate approach:

MAP OUT the distances in 30-minute increments until the police have traveled the same total distance as the gang.
Since the gang's rate = 50 mph, the gang travels 25 miles every 30 minutes.
Since the police's rate = 80 mph, the police travel 40 miles every 30 minutes.

Start --> gang = 50 miles, police = 0 miles
30 minutes later --> gang = 50+25 = 75 miles, police = 0+40 = 40 miles
1 hour later --> gang = 75+25 = 100 miles, police = 40+40 = 80 miles
1.5 hours later --> gang = 100+25 = 125 miles, police = 80+40 = 120 miles
2 hours later --> gang = 125+25 = 150 miles, police = 120+40 = 160 miles

Since the police are 5 miles behind after 1.5 hours (120 miles for the police versus 125 miles for the gang) but 10 miles ahead after 2 hours (160 miles for the police versus 150 miles for the gang), the time for the police to catch up must be between 1.5 hours and 2 hours.
Only C works.

To simplify:
The range between a police car and a train: 50 miles
The difference in 2 vehicles speed: 30 miles/hour
=> In 1 hour, the range reduces 30 miles. Then only 20 miles are left after 1 hour, obviously it does not need to take a full 1 hour to reduce 20 miles range. Only need: (20/30)* 60mins = 40 mins
The answer is: 1hour + 40 mins

Can anybody explain with picture/ diagram ?

This is a shrinking gap question.

Train's speed = 50 miles per hour
Police card's speed = 80 miles per hour
80 miles per hour - 50 miles per hour = 30 miles per hour
So, the gap between the train and the police car DECREASES at a rate of 30 miles per hour

Original gap (aka distance) = 50 miles
Time = distance/rate
So, time to close gap = 50/30 hours
= 5/3 hours
= 1 2/3 hours
= 1 hour and 40 minutes

Answer: C

Cheers,
Brent

gap between the train and the car 80-50=30 mph
diatance 50miles
time=distance/speed
=50/30 h = 5/3h = (5/3)*60= 100 minutes= 1 h 40 minutes

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