## Time Speed and Distance Tips and Tricks

Check out the formulas of Speed, Time and Distance here first if you have not already.

**When Distance is constant in the given scenario**

**Way to Solve type 0 Problem**

We all know that Distance = Speed x Time

So, if Distance if constant then this equation will be

S_{1} x T_{1} = S_{2} x T_{2}

Now, most of you will ignore this in the exam and it wouldn’t cross your mind to use this. Even if you understand it now.

But, if you make a habit of reading question properly not just for speed time and distance but for every topic you can solve the question in 1/4th of the time.

Same of this, we had done research on 8 engineering students solving a speed time distance problem only 2 used this formula and both solved the question correctly in 1/4th of the time of other 6 on average.

**Example of type 0 Problem**

A boy walking at a speed of 15 km/hr reaches his school 20 min late. Next time he increases his speed by 5 km/h but still he late by 5 min. Find the distance of the school from his home.

Let time be x.

So, quick method =>

- 15(x +20) = 20(x + 5)
- 300 – 100 = 5x
- x = 40. => x +20 = 60 mins = 1hr
- Thus distance = 15(1) = 15 kms

explanation to above

S_{1 }= 15km/h ; S_{2} = 5 + 15 = 20 km/h

T_{1} = x + 20 T_{2} = x + 5

Now just apply the formula

Now Same thing can be applied for Speed and Time also

**So the all three formula will be –**

**S**_{1}x T_{1}= S_{2}x T_{2}**D**_{1}/ T_{1}= D_{2}/ T_{2}**D**_{1}/ S_{1}= D_{2}/ S_{2}

**Object Travelling in Opposite Direction of the Train**

**Way 2 Solve Type 1 Problem**

If the object has negligible length then use the formula –

**Objects moving in Opposite directions**- (S
_{T }+ S_{O}) = L_{T}/ t

- (S
**Objects moving in Same directions**- (S
_{T }– S_{O}) = L_{T}/ t

- (S

If the Object has significant Length then use the formula

**Objects(Trains) moving in Opposite directions**- (S
_{T }+ S_{O}) = (L_{T }+L_{O})/ t

- (S
**Objects(Trains) moving in Same directions**- (S
_{T }– S_{O}) = (L_{T }+L_{O})/ t

- (S

**Example – **

Two trains are travelling in opposite directions at uniforms speeds of 60 kmph and 50 kmph. They take 5 seconds to cross each other.If the two trains travelled in the same directions.then a passenger sitting in the faster moving train would have overtaken the other than in 18 seconds.what are the lengths of the trains?

Let length of faster train and slower train be x and y respectively

__When they travel in opposite direction, trains cross each other in 5 sec__

Relative speed = 60+50 =110 km/hr = (110×5)/18= 550/18

Distance travelled = x+y

Time taken to cross each other = 5 s

Distance = Speed x Time

(x+y) = (550/18)×5=1375/9 –(1)

__When they travel in same direction, passenger sitting in the faster moving train would have overtaken the other train in 18 sec__

Relative speed = 60-50 =10 km/hr = (10×5)/18=50/18 m/s

Since the observation given is of a passenger sitting in faster train, distance travelled is equal to length of the slower train.

i.e, distance travelled = y m

Time taken for overtaking = 18 s

y = (50/18)×18 = 50 m

From(1), x + 50 = 13759

“>1375/9

x = 102.78 m

**Train A leaves at X am from Position 1 and Reaches Position 2 at Y am, other train leaves from Position 2 at A Am and Reaches at B AM. Distance is given. When do two Trains Cross one another.**

**Way to Solve type 2 Problem**

This is very frequently asked problem.

Let us understand this by an Example.

A train starts from Delhi at 6:00 am and reaches Ambala cantt. at 10am. The other train starts from Ambala cantt. at 8am and reached Delhi at 11:30 am, If the distance between Delhi and Ambala cantt is 200 km, then at what time did the two trains meet each other?

Average speed of train leaving Delhi = 2004=50

“>200/ 4 = 50

km/hr

Average speed of train leaving Ambala cantt. = 200×27=4007

“>200×2/7 =400/7

By the time the other train starts from Ambala cantt, the first train had travelled 100 km as 2 hours x 50kmph = 100 Kms

Therefore, the trains meet after:

=200−100(50+400/7)=1415

“>=(200−100)/(50+400/7)=14/15

hr

=1415×60=56

“>=(14/15)×60 = 56

minutes

Hence they meet at **8:56 am as 6 am + 2 hours + 56 mins**

**Had Object been Faster by a Km/hr then time taken and had it been slower then b Km/hr. What is the Distance?**

**Way to solve type 3 Problem**

Lets look this with an Example

A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hour less than the scheduled time. And, if the train were slower by 6 km/hr, the train would have taken 6 hr more than the scheduled time. The length of the journey is:

Let the length of the journey be d km and the speed of train be S

“>SS

km/hr.

Then,

dS+6=t−4

“>d/(S+6) = t−4

—–(i)

dS−6=t+6

“>d/(S−6) = t+6

——(ii)

Subtracting the 1 equation from another we get:

dS−6−dS+6=10

“>[d/(S−6)] − [d/(S+6)] = 10

——(iii)

Now t=ds

“>t = d/s

Substitute in equation (i) and solve for d

“>d

and S

“>S

**d = 720 km**

**A and B start walking towards one another at same time at given speeds and separated by a given distance. What is the time they cross one another at?**

**Way 2 Solve Type 4 Problem**

A will travel some distance d1 and B will travel d2. Sum of d1 and d2 will be D(given distance between the two) solve for this equation and time will be same for both.

- d1 = s1 x t
- d2 = s2 x t

Example.A and B start walking towards each other at 10 am at speeds of 3 km/hr and 4 km/hr respectively. They were initially 17.5 km apart. At what time do they meet?

Let after T

“>T

hours they meet

Then, 3T+4T=17.5

“>3T+4T = 17.5

T=2.5

“>T=2.5

Time = 10:00 am + 2.5 hour = **12:30 pm**