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, 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 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 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. Boats & Streams While moving in upstream, distance covered S= (U-V) T Incase of down stream, distance covered S= (U+V) T Circular Motion 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 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. QuestionS 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 Jim’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 |
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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
In the above eg: it is V1/[(3/4)*V1]
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