Speed Time and Distance Tips and Tricks

We all know that Distance = Speed x Time
So, if Distance if constant then this equation will be
S1 x T1 = S2 x T2
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 1 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.

Solution :

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
S1 = 15km/h ;
S2 = 5 + 15 = 20 km/h
T1 = x + 20 T2 = 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 –
S1 x T1 = S2 x T2
D1 / T1 = D2 / T2
D1 / S1 = D2 / S2

Object Travelling in Opposite Direction of the Train
Way 2 Solve Type 2 Problem
If the object has negligible length then use the formula –
Objects moving in Opposite directions
(ST + SO) = LT/ t
Objects moving in Same directions
(ST – SO) = LT/ t
If the Object has significant Length then use the formula
Objects(Trains) moving in Opposite directions
(ST + SO) = (LT +LO)/ t
Objects(Trains) moving in Same directions
(ST – SO) = (LT +LO)/ t

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?

Solution :

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
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 3 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?

Solution :

Average speed of train leaving Delhi = 200/4 = 50 km/hr
Average speed of train leaving Ambala cantt. = 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)=14/15 hr
=(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 4 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:

Solution :

Let the length of the journey be d km and the speed of train be S km/hr.
Then,
d/(S+6) = t−4 —–(i)
d/(S−6) = t+6 ——(ii)
Subtracting the 1 equation from another we get:
[d/(S−6)] − [d/(S+6)] = 10 ——(iii)
Now t = d/s
Substitute in equation (i) and solve for d and 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 5 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

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?

Solution :

Let after T hours they meet
Then, 3T+4T = 17.5
T=2.5
Time = 10:00 am + 2.5 hour = 12:30 pm