Speed, Time and Distance concepts with solved examples including shortcut method
Speed, Time, and Distance:
1.Speed: The distance covered in a unit time interval is known as speed.
Speed = Distance travelled/ Time taken
Distance = Speed * Time
Time = Distance/Spedd
Unit conversion:
i. Km/hr to m/sec.
A km/hr = A * (5/18) m/s
ii. m/sec to Km/hr.
A m/s = A * (18/5) Km/hr
how to remember: To reduce Unit from Km/hr to m/s we divide it by 18 i.e 5/18.
To increase the unit from m/s to km/hr we multiply it by 18 i.e 18/5.
Ex.1 A train is moving at a speed of 180 km/hr. Its speed in meters per second?
Sol.1 Km/hr = 5/18 m/sec
180 * (5/18) = 50 m/sec Ans.
Points to remember :
1. If the time taken is constant, then the distance traveled is directly proportional to the speed.
S ∝ D
2.If the speed is constant, then the distance traveled is directly proportional to the time taken.
D ∝ T
3.If the distance traveled is constant, the speed is inversely proportional to the time taken.
S ∝ 1/T
Ex.2 In covering a certain distance the speed of A and B are in the ratio of 3:4/ A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is?
Sol. Let the distance is D Km.
The speed of A is 3x km/hr
The speed of B is 4x km/hr
According to the question:
T1-T2 = 30min = 1/2
D/3x - D/4x = 0/2
D/12x = 1/2
D/x = 6
Time Taken by A = D/3x =6/3 = 2 hours.
Method 2.
Speed Ratio = 3:4
Time Ratio = 4:3
1 = 30 min
4 = 120 min = 2 hours
Ex.3 If a person walks at 14 km/hr instead of 10 km/hr. he would have walked 20 km more. The actual distance traveled by the person is?
Sol. Method 1
Let the actual distance traveled be x km.
Then,
Speed = Distance/ Time
x/20 = (x+20)/14
14x = 10x + 200
4x = 200
x = 50km
Method 2.
14-10 = 20/T
T =5 hrs
Distance = 10* 5 = 50 km Ans.
Average Speed:
The average speed of an object is a measure of the distance covered by that object in a set period of time.
Average speed = Total Distance /Total time taken
a. If A covers a distance d1 km at s1 km/hr and then d2 km at s2 km/hr. Then the average speed during the whole journey is
Average speed = s1s2(d1+d2)/(s1d1+s2d2)
b.A person goes a certain distance A to B at speed of s1 and returns B to A at speed of s2 km/hr. If he takes T hrs in all, then
Average speed = 2s1s2/(s1+s2)
Distance between A and B = t * [s1s2/(s1+s2)]
c.If a person traveled three equal Distance by three different Speed s1,s2,s3 then
Average speed = 3s1s2s3/(s1s2+s2s3+s3s1)
Ex4. A man goes from place A to B at a speed of 12 km/hr. and return at the speed of 18 km/hr. The average speed for the whole journey is.
Sol. Average speed = 2*12*18/(12+18)
2*12*18/30 = 72/5 km/hr
Ex5. One-third of a certain journey is covered at the rate of 25 km/hr/One-fourth at the rate of 30 km/hr. and the rest at 50 km/hr. The average speed for the whole journey.
Sol. Let the total journey = x km/hr
so according to the question
Total time = x/3*25 + x/4*30 + 5x/12*50
Average speed = x/(3x/100) = 100/3km/hr
Previous year questions of Time and Distance :
Ex.1 A student goes to school at a rate of 2.5 km/hr and reaches 6 minutes late. If he travels at a speed of 3 km/hr. he is 10 minutes early. What is the distance to the school?
Sol. Let the distance of the school be x km.
Diffrence of time = 6+10 = 16/60 hours = 4/15
Time = Distance/Speed
x/(5/2) - x/3 = 4/15
2x/75-x/3 = 4/15
x= 4 km
Method 2.
Required distance = (product of two speed/ difference of two speed) * Difference between the arrival time
[{(5/2)*3}/(3 - 5/2)] * (6+10)/60 = 4 km/hr
Ex. 2 The distance between 2 places R and S is 42 km. Anita starts from R with a uniform speed of 4 km/hr towards S and At some speed, Romita starts from S towards R. Also with some uniform speed they meet each other after 6 hours. The speed of Romita is?
Sol. Let the speed of Romita is x km/hr
Distance = speed * Time
The distance traveled by Romita and Anita together is 42 km.
4*6 + x*6 = 42
6x = 42-24
x = 3 km/hr Ans
Ex. 3.A is twice as fast as B and B is thrice as fast as C is. The journey covered by C in 1.5 hours will be covered by A in?
Sol. Let the speed of B, C, and A is x km/hr, 2x km/hr, x/3 km/hr.
Speed of A/Speed of B = 2x/(x/3) = 6
Required time = 1/6 * 3/2 hrs, = 1/4 hour = 15 minutes
Method 2
The ratio of speed A: B: C = 2: 1: 1/3
Multiply ratio by 3
Speed ratio A:B:C = 6:3:1
Time ratio A:B:C = 1:3:6
C covered journey in 1.5 hours means,
6 = 1.5 = 3/2
1 = 3/2 * 1/6 = 1/4 = 15 minutes
Ex.4 A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr if he takes 5 hours in going and coming. The distance between his house and school?
Sol. Let the distance is x km
then
x/3 + x/2 = 5
(2x+3x)/6 = 5
x= 6 km
Method 2.
Distance = total time taken * (product of two speed)/ addition of two speed
5 + (3*2)/(3+2) = 6 km
Ex.5 A man traveled a distance of 50 km in 7 hrs partly on foot at the rate of 8 km/hr. and partly on bicycle at 16 km/hr. The distance traveled on the foot?
Sol. Journey on foot = x km
Journey on cycle = 80 - x km
x/8 + (80-x)/16 = 7
On solving the above equation we get.
x= 32 km
Ex.6 Walking 6/7 of his usual speed a man is 12 minutes late the usual time taken by him to cover that distance is?
Sol. Time and speed are inversely proportional.
Therefore,
Usual time * 7/6 - usual times = 12 minutes
Usual time * 1/6 = 12 minutes
Usual time = 72 minutes = 1 hour 12 minutes
Method 2.
Usual time = late time / {1/(6/7) -1}
= 12/{(7/6) - 1} = 12/(1/6) = 72 minutes = 1 hour 12 minutes
Ex.7 Running 4th of his usual speed, a person improves his timing by 10 min. Fin the usual time to cover the distance?
Sol. Usual time = improved time / {1- 1/(4/3)}
10/(1-3/4) = 40 minutes Ans.
Ex. 8 A car covers four successive 7 km distances at speeds 10 km/hr, 20 km/hr, 30 km/hr, and 60 km/hr respectively. Its average speed over this distance is.
Sol. Total distance = 7*4 = 28 km
Total time = 7/10 +7/20 + 7/30 + 7/60
(42+21+17+7)/60 = 84/60 = 7/5
Average speed = total distance / total time = 28/(7/5) = 20 km/hr Ans.
For question-based on relative speed and trains click here.
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