# Time , Speed and Distance

We define speed as distance divided by time,

Speed = distance /time (D=ST)

but once we have the equation, we can use any of its variations,

speed = distance / time ,

distance = speed * time and

time = distance / speed to compute any one of the quantities when we happen to know the other two. For example, suppose we drive for 2 hours at 30 miles per hour, for a total of 60 miles. If we know the time and the speed, we can find the distance: 2 hours * 30 miles/hour = 60 miles If we know the time and the distance, we can find the speed: 60 miles / 2 hours = 30 miles/hour

Distance is directly proportional to Velocity when time is constant

1. A car travels at 30km/hr for the first 2 hrs & then 40km/h for the next 2hrs. Find the ratio of distance traveled

S1/S2=V1/V2=3/4

2. Two cars leave simultaneously from points A & B (100km apart)& they meet at a point 40 km from A. What is Va/Vb?

T is constant so V1/V2=S1/S2=40/60=4/6

3. A train meets with an accidient and moves at 3/4th its original speed. Due to this , it is 20 mins late. Find the original time for the journey beyond the point of accident?

Method1 : Think about 2 diff. Situations , 1st with accident and another w/o accident , then S is constant

So V1/V2=T1/T2 =>V1/[3/4)*V2]=(T1+20)/T1

=>4/3=(T1+20)/T1 =>T1=60

Method 2: Velocity decreases by 25% so time will increase by 33.3%

33.3%=20 mins =>100%=60 mins

**CONVERSION:** 1km/hr=1000m/h=1000/3600m/sec=5/18m/sec

RELATIVE SPEED

**Caes1:** Two bodies are moving in opposite directions at speed V1 & V2 respectively. The relative speed is defined as Vr = V1 + V2

**Case2:** Two bodies are moving in same directions at speed V1 & V2 respectively.The relative speed is defined as Vr = |V1 - V2|

Train Problems

The basic equation in train problem is the same S=VT

The following things need to be kept in mind while solving the train related problems.

When the train is crossing a moving object , the speed has to be taken as the relative speed of the train with respect to the object.

The distance to be covered when crossing an object, whenever trains crosses an object will be equal to: Length of the train + Length of the object

Boats & Streams

Let U= Velocity of the boat in still water

V=Velocity of the starem.

While moving in upstream, distance covered S= (U-V) T

Incase of down stream, distance covered S= (U+V) T

Circular Motion

The relative velocity of 2 bodies moving around a circle in the same direction is taken as (V1 â€“ V2) and while moving in opposite direction is taken as (V1+V2)

First Meeting Three or more bodies start moving simultaneously from the same point on the circumference of the circle . They will 1st meet again in the LCM of the times that the fastest runner will takes in totally overlapping each of the slower runners

First meeting time = Circumference / Relative velocity

First Meeting at starting point The first meeting at the starting point will occur after a time that is obtained by the LCM of the times that each of the bodies takes the complete one full round.

CLOCK

For clock problems consider the clock as a circular track of 60km.

Min. hand moves at the speed of 60km/hr (think min. hand as a point on the track) and hour hand moves at 5km/hr and second hand at the speed of 3600 km/hr.

Relative speed between hr hand and mins hand = 55

**QuestionS**

1.A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

The correct answer is 240 meters.

2. A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds. What is the length of the train in meters? Correct answer is 400 meters.

3. Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. what is the average speed of Jimâ€™s travel in mph? Correct answer is 37.5 mph.

4. A runs 25% faster than B and is able to give him a start of 7 meters to end a race in dead heat. What is the length of the race? The correct answer is 35 meters.

5. Jane covered a distance of 340 miles between city A and city taking a total of 5 hours. If part of the distance was covered at 60 miles per hour speed and the balance at 80 miles per hour speed, how many hours did she travel at 60 miles per hour? The correct answer is 3 hours.

6. Steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 80 mph. What is his average speed of travel for the entire journey? The correct answer is 64 mph

So V1/V2=T1/T2 =>** V1/[3/4)*V2**]=(T1+20)/T1

=>4/3=(T1+20)/T1 =>T1=60

you made a mistake the correct ans is

V1/v2=T1/T2 => V1/[(3/4)*V1 ]= T1/(T1+20)

4/3= T1/(T1+20)

=> T1=80

So V1/V2=T1/T2 =>** V1/[3/4)*V2**]=(T1+20)/T1

=>4/3=(T1+20)/T1 =>T1=60

you made a mistake the correct ans is

V1/v2=T1/T2 => V1/[(3/4)*V1 ]= T1/(T1+20)

4/3= T1/(T1+20)

=> T1=80

they are inversely proprotional ie speed and time

problem :

If Arun had walked 1 km/h faster, he would have taken 10 minutes less to walk 2 kilometres. What is Arun's speed of walking?

no the ans is not 80 it is 60

see v1/v2=t2/t1

v1/((3/4)v1)=(t1+20)/t1

after solving this u will get t1=60

is the ques to ur answer is 2km/hr or 3 km/hr .........kindly reply, the ques has stormed my mind

speed can not be constant since accident happens or nt distance will be the same.

so keeping distance cons, we have speed inversely proport. to time.

so when speed is 4:3

time will be 3:4

and the multiplying factor will be 20(as given)

therefore.

t1= 3*20=60

and t2= 4*20=80.

3. A train meets with an accidient and moves at 3/4th its original speed. Due to this , it is 20 mins late. Find the original time for the journey beyond the point of accident?

Method1 : Think about 2 diff. Situations , 1st with accident and another w/o accident , then S is constant

So V1/V2=T1/T2 =>

]=(T1+20)/T1V1/[3/4)*V2=>4/3=(T1+20)/T1 =>T1=60

In the above eg: it is V1/[(3/4)*V1]