Help with distance catch up problem?

You know the distance formula is d = rt.

Let t be the usual time it takes (in minutes)
Let r be its usual speed
d = rt

Then when it is going slower:
The time it takes is t + 30.
The rate is r/3.
d = (t + 30)r/3

Equate these two:
rt = (t + 30)r/3
rt = rt/3 + 30r/3
rt = rt/3 + 10r
rt - rt/3 = 10r
rt(1 - 1/3) = 10r
rt(2/3) = 10r

Divide both sides by r (assuming it is non-zero):
(2/3)t = 10
t = 10 * 3/2
t = 30/2
t = 15

That's a longer, more algebraic way to solve this.

But here's an intuitive way to solve it.

The speed and the time are inversely proportional. In other words, if the speed is 1/3 the usual speed, the time will be 3 times as much as usual.

Let t be the usual time.
Let 3t be the new time.
3t = t + 30
2t = 30
t = 30/2
t = 15 minutes

Summary:
Normally it takes 15 minutes, but when it goes 1/3 as fast it will take 3 times as long (45 minutes) which is 30 minutes longer.

Answer:
15 minutes

Let t min be the usual time to cover the same distance.

When velocity = 3v units/min, time taken = t min:
Hence, distance = 3v • t units ... [1]

When velocity = (1/3)(3v) = v units/min, time taken (t + 30 min):
Hence, distance = v • (t + 30) units ... [2]

[1] = [2]:
3v • t = v • (t + 30)
3t = t + 30
2t = 30
t = 15

Usual time to cover the same distance = 15 min

distance = speed * time.  Let  V be the speed and T be the time
distance = V * T
now with the new information remembering that the distance is the same no matter how fast or slow you travel at
distance = V / 3 * (T + 30)
setting the two equations equal to each other because distance is the same gives you
V * T = V/3 * (T + 30)
VT = VT/3 + 10V
2/3VT = 10V
T = 15 minutes or 0.25 hour
 
Try some numbers and see if it works. Say the speed = 60 miles/hr.  Distance = 15 miles.  If the speed is dropped to 20 mile/hr then it takes 0.75 hours, 0.5 hr or 30 minutes longer.
Try 10 miles/hr.  Distance = 10 mile/hr * 0.25 hr = 2.5 miles.   Drop the speed to 1/3 and it now takes 0.75 hours.  Again, 30 minutes faster.

Recall: s = d/t → where s is the speed, d is the distance, t is the time


First case: usual speed

s = d/t → where d is the distance to reach the destination

s = d/t → where t is the required time to reach the destination at normal speed


Second case: usual speed divided by 3, time is 30 minutes more, i.e. (1/2) hours

s = d/t → adapt this formula to the present case

s/3 = d/[t + (1/2)] → recall: s = d/t

(d/t)/3 = d/[t + (1/2)]

d/(3t) = d/[t + (1/2)] → you can simplify by d both sides

1/(3t) = 1/[t + (1/2)]

3t = t + (1/2)

2t = 1/2

t = 1/4 ← in hours

t = 15 minutes

 usual time to cover the same distance is 15 minutes.

Step-by-step explanation:

Required Formula:

Time = Distance / Speed

Let's assume

The usual speed of the train be "s" km/hr

The distance travelled by the train be "D" km.

The usual time taken by the train to cover the same distance be "t" hr = D/s
Given that,

The reduced speed of the train = km/hr

Therefore, according to the question, we can write the eq.
 
- D/(s/3) -D/s=30min/60min

3D/s-D/s=1/2

D/s(3-1)=1/2

D/s(2)=1/2

D/s=1/4



Thus,
The usual time "t" taken by the to cover the same distance is,
= D/s
= 1/4*60min
= 15 minutes

Let s = Usual speed in mph
Let s/3 = Speed now in mph

Let t = Usual time in hours
Let t + 1/2 = Time now in hours

Let d = Distance in miles.

Then.

t = d / s. This is the "usual" situation.

t + 1/2 = d / (s / 3). This is the current situation.

We can now solve these equations for t.

t = d / s

2t + 1 = 2d / (s / 3) = 6d / s = 6t

4t = 1

t = 1/4 hour or 15 minutes.