8. Speed, Time and Distance(examveda PDF

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Mathematics (From www.examveda.com) i.e. speed α distance. (When time is constant) A body travels at S2 kmph for the first 2 hours and then travels at S2 kmph for the next two Speed, Time and Distance and Its Applications hours. Concept of speed, time and distance is based on In such case the following ratios will be valid: S2/S2 = Da/Db the formula 3. When distance is constant, speed is inversely Speed * time = distance proportional to time. Speed (s): Speed can be defined as rate at which distance is i.e. speed α 1/time. In such case, the following ratios will be valid: covered during the motion. It is measured in S1/S2 = T2 / T1 terms of distance per unit time. (i.e. m/s, Illustrations: km/hour, m/min, km/min, etc.). A train meets with an accident and moves at 3/4 Speed = distance/time of its original speed. Due to this, it is 20 minutes Time (t): It is the time duration over which the movement late. Find the original time for the journey has occurred. The unit used for measuring time beyond the point of accident. Solution: is synchronous with denominator of the unit used for measuring speed. Thus, if the speed is In the above case, distance of whole journey is constant. measured in terms of km/h then time is Then, we can use, speed a 1/time. measured in hours. Due to accident speed become 3/4 then time Time = distance/ Speed becomes 4/3. Distance (d): Since, extra time = 1/3 of normal time= 20 It is the displacement of the body during the minutes. motion. Hence, normal time= 60 minutes. Distance= Speed * time

Speed, Time and Distance

Key Points of the equation, Speed * time= distance. 1. When speed is constant, time is directly proportional to distance. Time α distance (when speed is constant.) Illustration: A car moves for 2 hours at a speed of 25 kmph and another car moves for 3 hours at the S2me speed. Find the ratio of distances covered by two cars. Since, the speed is constant, we can conclude that time α distance. Hence, Ta /Tb = Da /Db Since, the times traveled are 2 and 3 hours respectively; the ratio of distance covered is also 2/3.

Conversion between kmph to m/s: 1 km/h= 1000 m/s= 1000/3600 m/s =5/18 m/s. Hence, to convert y km/h into m/s multiply by 5/18. Thus, y km/h= 5y/18 m/s. And vice-versa: y m/s= 18y/5 km/h. To convert from m/s to kmph, multiply by 18/5. Example 1: A train is running with the speed of 45 km /h. What its speed in meter per second? Solution: Speed of train in meter/s= 45*5/18=12.5 m/s. Example 2: If a motor car covers a distance of 250 m in 25 seconds, what is its speed in kilometer per hour? Solution: Speed of motor car= 250*18/5= 36 km/h.

2. When time is constant speed is directly proportional to distance. Published by Exam Aid Publication

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Mathematics (From www.examveda.com) Concept of Relative Speed Movement of one body with respect to another moving body is called the relative speed of each other. Case 1: Two bodies are moving in opposite direction at a speed S1 and S2 respectively. Then Relative speed = S1 +S2 Case 2: Two bodies are moving in same direction at a speed S1 and S2 respectively. Then Relative speed= |S1 -S2| In this case, we take positive value of difference of speeds.

Example 2: A car travels at 60 km/h from Mumbai to Poona and at 120 km/h from Poona to Mumbai. What is the average speed of the car for the entire journey?< Solution: Average speed= 2S1 *S2/ (S1 +S2) Average speed= 2*60*120/180 Average speed= 80 km/h.

Acceleration: Acceleration is defined as rate of change of speed. Acceleration can be positive (speed increases) or negative (speed decreases, also known as deceleration). Unit of acceleration is m/S2 Final speed= Initial speed + Acceleration * time.

These points need to be kept in mind while solving a train related questions.

Average speed: If a certain distance is covered in parts at different speeds, the average speed is given by Average speed = Total distance covered / total time taken. Suppose a car goes from A to B at an average speed S1 and then comes back from B to A an average speed of S2. If you have to find the average speed of the whole journey, then average speed is given by, Average speed = 2S1 *S2/ (S1 +S2) Average speed= 2* products of speeds/ sum of speeds. Example 1: Ramu climbs a mountain with 10 km/h speed and returns with 20 km/h speed then what is his average speed? Solution: Average speed= 2S1 *S2/ (S1 +S2) = 2*10*20/10+20=13(1/3) km/h.

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Trains Trains are special case in the concept of speed, time and distance. The basic relation is same as, Speed * time= distance.

1) When a train is crossing a moving object, the speed has to be taken as relative speed of the train with respect to the object. 2) The distance to be covered while crossing an object whenever a train crosses an object will be equal to Length of train + Length of the object 3) Time taken by the train to cross a pole, man etc. by a train, = Length of train /Speed of train. Example: A train crosses a pole in 8 seconds. If the length of the train is 200 meters, find the speed of train. Solution: Time taken to cross the pole= Length of train/Speed of train. Speed of train = Length of train/ time taken to cross the pole. Speed of train=200/8=25 m/s. = 25*18/5= 90 km/h. 1) Time taken to cross another train or bridge or platform etc. by a train is given by, = Length of (train + platform etc.)/Speed of the train. Page 145

Mathematics (From www.examveda.com) Example: A 150 m long train passes another train 100 m long traveling in opposite direction. If the speeds of the trains are 39 km/h and 21 km/h respectively, how long will it take to cross each other? Solution: Total distance= length of 1st train + length of 2nd train = 150+100= 250 m. Since, trains are moving in opposite directions, their relative speed= 39+21= 60 km/h. = 60*5/18=50/3 m/sec. Hence, time taken to cross= 250/(50/3) = 250*/50= 15 seconds.

Example: A monkey tries to climb a pole 12 m high. In each minute he climbs 2 m but slips down 1 m. how much time will be required by him to reach the top? Solution: Reqd. time= [(ht. of pole - slipped distance)/(climbed distance - slipped distance)]*t. = [(x-z)/(y-z)]*t = [12-1/2-1]*1= 11 minute. Boat and Stream The problem of boats and streams are also based on formula, Speed * time= distance

Key Facts: 1) If two persons start at the same time from two different positions towards each other and they complete their journey in 'a' and 'b' hours respectively after meeting each other then(Speed of A/ Speed of B)= sqrt of (b/a). Example: Two trains are approaching each other from two different points at the same time. They complete their journey in 3(1/3) and 4(4/5) hours after they met each other. Speed of A is 8 km/h. what is that of speed of B. Solution: (Speed of A/ Speed of B)=Sqrt of [4(4/5)] /sqrt of [3(1/3)]. 8/B= sqrt of (24/5)/sqrt of (10/3) On squaring both sides, we get, 64/B2=24*3/10*5. B2 =64*10*5/24*3=400/9 B = 20/3=6(2/3) 1) A monkey or man tries to climb a pole 'x' m high. In 't' time he climbs 'y' m but slips down 'z' m. How much time will be required by him to reach the top? Reqd. time = [(ht. of pole - slipped distance)/(climbed distance - slipped distance)]*t = [(x-z)/(y-z)]*t Published by Exam Aid Publication

Some points need to be kept in mind while solving problem related with Boat and stream. 1) Speed of boat moving downstream or in direction of flow of stream = speed of boat in still water+ speed of stream. 2) Speed of boat moving upstream or in opposite direction of the flow of stream = Speed of boat in still water - speed of stream , 3) Speed of boat in still water = speed of boat Circular Motion When two or more bodies are moving around a circular track, the motion around this circular track is known as circular motion. The relative motion of two bodies moving around a circle in the same direction is taken as = S1 - S2 Also, when bodies are moving in opposite direction, then relative speed = S1 +S2

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Mathematics (From www.examveda.com) Clocks Problems on clocks are based on the movement of the minute hand and that of the hour hand as well as on the relative movement between two. For the simplicity, we don't take the second hand. It is best to solve problems of clocks by considering a clock to be in a circular track having a circumference of 60 km and each kilometer being represented by one minute on the dial of the clock. Then, we can look at the minute hand as runner running at the speed of 60kmph while we can also look at the hour hand as a runner running at an average 5kmph (60/12=5kmph). Since, minute hand and hour hand are running in same direction, the relative speed of minute hand with respect to hour hand is 55kmph (60-5=55kmph). This means that for every hour elapsed, the minute hand goes 55 km (minute) more than the hour hand.

when the minute hand is 15 kilometers ahead of the hour hand. Thus, the minute hand has to cover 25 kilometer over hour hand. Distance covered by minute hand over hour hand; 55 kilometer _________ in 1 hour. Then, 1 km ______________ in 1/55 hour. Hence, 25 km ___________ in 25/55= 5/11 hour. Thus, the first right angle between 2-3 is formed at 5/11 hours past 2 o'clock. 5/11= 5*60/11=27 (3/11) minutes. 3/11 minute= 3*60/11= 16.3636 second. Hence, the required answer is; 2:27:16.36 pm.

Key facts: ==> A clock makes two right angles between any 2 hours. ==> Right angles are formed when the distance between the minute hand and hour hand is equal to 15 minute. ==> A clock makes two straight lines between any 2 hours. Straight lines are formed when the distance between the minute hand and the hour hand is equal to either 0 minutes or 30 minutes. Illustration: At what time between 2-3 pm is first right angle formed by hands of clocks? Solution: At 2 pm minute hand can be visualized as being 10 kilometers behind the hour hand as the minute hand is 10 minute behind the hour hand. The first right angle between 2-3 is formed Published by Exam Aid Publication

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Mathematics (From www.examveda.com) 2. Waking 3/4 of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between 1. Two buses start from a bus terminal with a his home and office: speed of 20 km/h at interval of 10 minutes. A. 48 min. B. 60 min. What is the speed of a man coming from the C. 42 min. D. 62 min. opposite direction towards the bus terminal if he E. 66 min meets the buses at interval of 8 minutes? Answer: Option A A. 3 km/h B. 4 km/h Solution: 1st method: C. 5 km/h D. 7 km/h 4/3 of usual time = Usual time + 16 minutes; Answer: Option C Hence, 1/3rd of usual time = 16 minutes; Solution: Let Speed of the man is x kmph. Thus, Usual time = 16*3 = 48 minutes. Distance covered in 10 minutes at 20 kmph = 2nd method: distance covered in 8 minutes at (20+x) kmph. When speed goes down to Or, 20*10/60 = 8/60(20+x) 3/4th (i.e. 75%) time will go up to 4/3rd (or Or, 200 = 160+8x 133.33%) of the original time. Or, 8x = 40 Since, the extra time required is 16 minutes; it Hence, x = 5kmph. should be equated to 1/3rd of the normal time. Hence, the usual time required will be 48 Detailed Explanation: minutes. A _____________M_______________B A = Bus Terminal. 3. Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from B = Let meeting point of first bus and the man Delhi will the two trains be together: and this distance is covered by Bus in 10 A. 262.4 km minutes. I.e. Distance A to be is covered first B. 260 km bus in 10 min. As AB distance can be covered C. 283.33 km by second bus in 10 minutes as well. D. 275 km Distance Covered by Bus in 10 min = AB = E. None of these (20/60) * 10 = 10/3 km. Answer: Option C Now, M is the Meeting Point of Second Bus Solution: Difference in time of departure with Man. Man covered distance B to M in 8 between two trains = 45 min. = 45/60 hour = 3/4 minutes. hour. Now, Relative distance of both Man and Bus Let the distance be x km from Delhi where the will be same as both are traveling in opposite two trains will be together. direction of each other. Let Speed of the man = Time taken to cover x km with speed 136 kmph be t hour and time taken to cover x km with x kmph. speed 100 kmph (As the train take 45 mins. Relative speed = 20 +x more) be To meet at Point M, bus and Man has covered (t+3/4) = ((4t+3)/4); the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus Now, 100*(4t+3)/4 = 136t; in 10 minutes. Thus, Distance covered in 8 Or, 25(4t+3) = 136t; minutes with relative speed (20 +x) kmph = Or, 100t+75 = 136t; distance covered by bus in 10 mintes with speed Or, 36t = 75; 20 kmph. Or, t = 75/36 = 2.08 hours; Then, distance x km = 136*2.084 = 283.33 km. Section 1

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Mathematics (From www.examveda.com) 6. From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hour. If A traveled with 2/3 of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is: A. 4 km/h B. 6km/h C. 10 km/h D. 12 km/h Answer: Option B Solution: A →_______60Km_________← B Let the speed of A= x kmph and that of B = y Time taken in walking one way and riding back kmph; = 6 hours 15 minutes ----------- (ii) According to the question; X*6+y*6 = 60 By the equation (ii)*2 - (i), we have, Or, x+y = 10 --------- (i) Time taken by the man in riding both ways, And, {(2x/3)*5}+(2y*5) = 60 = 12 hours 30 minutes - 7 hours 45 minutes Or, 10x+30y = 180; = 4 hours 45 minutes. Or, x+3y = 18; ---------- (ii) 5. A man completes a certain journey by a car. If he covered 30% of the distance at the speed of From equation (i)*3 - (ii) 20kmph. 60% of the distance at 40km/h and the 3x+3y-x-3y = 30-18 Or, 2x = 12 remaining of the distance at 10 kmph, his Hence, x = 6 kmph. average speed is: A. 25 km/h 7. A, B and C start together from the same place B. 28 km/h to walk round a circular path of length 12km. A C. 30 km/h walks at the rate of 4 km/h, B 3 km/h and C 3/2 D. 33 km/h km/h. They will meet together at the starting Answer: Option A place at the end of: Solution: A. 10 hours Let B. 12 hours the total distance be 100 km. C. 15 hours Average speed = total distance covered/ time D. 24 hours taken; Answer: Option D = 100/[(30/20)+(60/40)+(10/10)]; Solution: Time taken to complete the revolution: = 100/[(3/2)+(3/2)+(1)]; A→12/4 = 3 hours; = 100/[(3+3+2)/2] B→ 12/3 = 4hours; = (100*2)/8 C→12*2/3 = 8 hours; = 25 kmph. Required time, = LCM of 3, 4, 8. = 24 hours. 4. A man takes 6 hours 15 minutes in walking a distance and riding back to starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride back both ways is: A. 4 hours B. 4 hours 3o min. C. 4 hours 45 min. D. 5 hours Answer: Option C Solution: Time taken in walking both the ways = 7 hours 45 minutes -------- (i)

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Mathematics (From www.examveda.com) 8. Ravi and Ajay start simultaneously from a place A towards B 60 km apart. Ravi's speed is 4km/h less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a places 12 km away from B. Ravi's speed is: A. 12 km/h B. 10 km/h C. 8 km/h D. 6 km/h Answer: Option C Solution: Ajay → (x+4) kmph. A ________ 60 km _________ B Ravi → x kmph. Let the speed of Ravi be x kmph; Hence, Ajay's speed = (x+4) kmph; Distance covered by Ajay = 60 + 12 = 72 km; Distance covered by Ravi = 60 - 12 = 48 km. According to question, 72/(x+4) = 48/x Or, 3/(x+4) = 2/x 3x = 2x+8 Or, x = 8 kmph. 9. The speed of A and B are in the ratio 3:4. A takes 20 minutes more than B to reach a destination. Time in which A reach the destination? A. 1(1/3) hours B. 2 hours C. 2(2/3) hours D. 1(2/3) hours Answer: Option A Solution: Ratio of speed = 3:4; Ratio of time taken = 4:3 (As Speed ∝ (1/Time), When distance remains constant.) Let time taken by A and B be 4x and 3x hour respectively. Then, 4x-3x = 20/60; Or, x = 1/3; Hence, time taken by A = 4x hours = 4*1/3 = 1(1/3) hours. Published by Exam Aid Publication

10. A man covers half of his journey at 6km/h and the remaining half at 3 km/h. His average speed is A. 9 km/h B. 4.5 km/h C. 4 km/h D. 3 km/h Answer: Option C Solution: Average speed = 2xy/(x+y); = 2*6*3/(6+3); = 36/9 = 4 kmph. 11. Two guns are fired from the same place at an interval of 6 minutes. A person approaching the place observes that 5 minutes 52 seconds have elapsed between the hearings of the sound of the two guns. If the velocity of the sound is 330 m/sec, the man was approaching that place at what speed (in km/h)? A. 24 B. 27 C. 30 D. 36 Answer: Option B Solution: Difference of time = 6 min-5 mins. 52 secs = 8 secs. Distance covered by man in 5 mins. 52 secs = distance covered by sound in 8 secs = 330*8 = 2640m. Hence, speed of man = 2640/5 min. 52 secs; = 2640m/352 secs = 2640*18/352*5 kmph = 27 kmph. 12. Running at 5/4 of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is: A. 30 min. B. 28 min. C. 25 min. D. 23 min. Answer: Option C Solution: When the athlete walks at 5/4 of his usual speed then he takes 4/5 of his usual time (As Speed ∝ (1/Time)). and he saves 5 minutes. → Usual time - (4/5)*usual time = 5 minutes; →( 1/5)* Usual time = 5 minutes; Usual time = 25 minutes.

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Mathematics (From www.examveda.com) 13. In a race of 1000 m, A can beat B by 100m. In a 400m, B beats C by 40 m. In a race of 500m. A will beat C by A. 95 m B. 50 m C. 45 m D. 60 m Answer: Option A Solution: When A runs 1000 m, B runs 900 m. Hence, when A runs 500 m, B runs 450 m. Again, when B runs 400 m, C runs 360 m. And, when B runs 450 m, C runs = 360*450/400 = 405 m. Required distance = 500-405 = 95 meter. That means when A runs 500 meter then B can run 450m then C runs 405 m. 14. A runs twice as fast as B and B runs thrice as fast as C. The distance covered by C in 72 minutes, will be covered by A in: A. 18 minutes B. 24 minutes C. 16 minutes D. 12 minutes Answer: Option D Solution: Ratio of the speed of A, B and C = 6:3:1 Then, ratio of time taken; = 1/6:1/3:1 = 1:2:6; Hence, time taken by A = 72/6 = 12 minutes.

Speed of the motorboat up-stream, = Distance /time taken = 56 km /(7/4)h = 56*4/7 = 32 kmph Let the speed of the current = x kmph Hence, 36-x = 32 Or, x = 36-32 = 4 kmph Speed of boat down the stream = 36+4 = 40 kmph. 16. An athlete runs 200 meters race in 24 seconds. His speed in km/h is A. 20 B. 24 C. 28.5 D. 30 Answer: Option D Solution: Speed of athlete, = 200m/24secs = 200*18/5*24 = 30 km/hour.

17. If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 5 minutes late. If he travels at the speed of 4 km/h, he reaches the school 5 minutes earlier than school time. The distance of his school from his house is: A. 1 km B. 2 km C. 3 km D. 4 km Answer: Option B 15. A motorboat in still water travels at speed of Solution: Let the distance between school and home be x km. 36 kmph. It goes 56 km upstream in 1 hour 45 minutes. The time taken by it to cover the same The difference of time when A goes school to school with these two different speed is 10 min distance down the stream will be: = 10/60 hour. A. 2 h, 25 min. B. 3 h Now, C. 1 h, 24 min. (x/3)-(x/4)= 10/60 D. 2 h, 21 min. Or, x/12 = 1/6 Answer: Option C Or, x = 12/6 Solution: 1 hour 45 minutes = 1+(45/60) = 7/4 = 2 km. hours.

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Mathematics (From www.examveda.com) 18. A man starts climbing a 11 m high wall at 5 pm. In each minute he climbs up 1 m but slips down 50 cm. At what time will he climb the wall? A. 5:30 pm B. 5:21 pm C. 5:25 pm D. 5:27 pm Answer: Option B Solution: 1st Method: Man climbs 1 m and slips down 50 cm (0.5 m) in one minute i.e. he climbs (1-0.5 = 0.5 m) in one minute. But in the last minute he will be climbing 1 m as he gets on the top so no slip. T...