Formulas:
Speed,Time and Distance:Speed = $\left(\dfrac{distance}{time}\right)$ , Time = $\left(\dfrac{distance}{speed}\right)$ , Distance = $ \left(speed \times time\right)$
km/hr to m/sec conversion:x km/hr = $\left(x\times\dfrac{5}{18}\right)$ m/sec.
m/sec to km/hr conversion:x km/hr = $\left(x\times\dfrac{18}{5}\right)$ km/hr
If the ratio of the speeds of A and B is a : b, then the ratio of the the times taken by then to cover the same distance is $\dfrac{1}{5}:\dfrac{1}{5}$ or b:a
Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then, the average speed during the whole journey is $\left(\dfrac{2xy}{x+y}\right)$ km/hr.
Distance is a numerical description of how far apart two objects are.
Formula
Distance = $\left(speed \times time\right)$
Example
A bus moves at a speed of 80 km in 5 hrs. Find the Distance?
here,
Speed of Bus is = 80 km
Time = 5 hrs
Distance = $\left(speed \times time\right)$
= $(80 \times 5)$
= 400 km
Exercise
What is the distance travelled on foot?
Let the time in which he travelled on foot = x hr
Then the time in which he travelled on bicycle = (9−x) hr
distance = speed $ \times$ time
=> 4x + 9(9 − x) = 61
=> 4x + 81− 9x = 61
=> 5x = 20
=> x = 4
Distance travelled on foot = 4x = 4 $ \times$ 4 = 16 km
Conversion of km/hr to m/sec and m/sec to km/hr
x km/hr = $\left(x\times\dfrac{5}{18}\right)$ m/sec.
m/sec to km/hr conversion
x m/sec = $\left(x\times\dfrac{18}{5}\right)$ km/hr
Example
Express a speed of 36 kmph in meters per second?
36 * 5/18 = 10 mps
Exercise
Relation between Distance, Speed and Time
Speed and time are inversely proportional (when distance is constant)
Which implies ->
Time = $\left(\dfrac{distance}{speed}\right)$
Example
A car covers a distance of 150 km in 3 hrs. Find the speed of the car?
Speed = $\left(\dfrac{distance}{time}\right)$
= $\dfrac{150}{3}$
= 50 kmph
Exercise
Solution: Speed = 10 km/hour
Distance covered = 20 km
Time = distance/speed
= 20/10 hour
= 2 hours
Note
If the ratio of the speeds of A and B is a:b, then, the ratio of the time taken by them to cover the same distance is
Formula
Example
The Speed of A and B is 2 km/hr and 3 km/hr , then the time taken by them to cover the same distance is
here,
Speed of A = 2 km/hr
Speed of B = 3 km/hr
$\dfrac{1}{a}: \dfrac{1}{b}$ = b : a
$\dfrac{1}{2}: \dfrac{1}{3}$ = 3 : 2
Exercise
If he takes 5 hours in going and coming, what is the distance between his house and office?
Speed of House to Office is = 3 km/hr.
Speed of Office to House is = 2 km/hr.
Ratio of his speed = 3 : 2
Therefore, ratio of the time taken = 2 : 3
Since total time taken is 5 hours, he has taken 2 hours to travel to his office and 3 hours to come back.
Distance between his house and office
= 2 $ \times$ 3 = 6 km
Average Speed
If an object covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey is
Example
A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr.
What is the average speed for the first 320 km of the tour?
Average Speed = $\dfrac{2 \times 64 \times 80}{64+80}$
=$ \dfrac{2 \times 64 \times 80}{144} $
= $\dfrac{2 \times 32 \times 10}{9}$
= 71.11 kmph
Exercise
and second half at the rate of 24 km/hr. Find the total journey in km.
Average Speed = $ \dfrac{2 \times 21 \times 24}{21+24}$ = 22.4 km/hr
Total distance = 22.4 $\times $ 10 = 224 km
Distance = $(240 \times 5)$ = 1200 km.
Speed = Distance/Time
Speed = 1200/$\dfrac{5}{3}$ km/hr. [We can write 1$\dfrac{2}{3}$ hours as 5/3 hours]
| $\therefore$ Required speed =$ \left(1200 \times\dfrac{3}{5} \right) $km/hr= 720 km/hr. |
Suppose A, B and C take $ x $,$\dfrac{x}{2}$ and $\frac{x}{3}$ days respectively to finish the work.
Then,$ \left(\dfrac{1}{x} +\dfrac{2}{x} +\dfrac{3}{x} \right) $=$ \dfrac{1}{2} $
$\Rightarrow \dfrac{6}{x} $=$ \dfrac{1}{2} $
$\Rightarrow x $ = 12.
So, B takes $\left(12/2\right)$ = 6 days to finish the work.
Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes $ x $ days to do the work.
Then, 10 : 13 :: 23 : $ x $
x =$ \left(\dfrac{23 \times 13}{10} \right) $
x=$ \dfrac{299}{10} $.
As 1 days work =$ \dfrac{1}{23} $;
Bs 1 days work =$ \dfrac{10}{299} $.
$\left(A + B\right)$s 1 days work =$ \left(\dfrac{1}{23} +\dfrac{10}{299} \right) $=$ \dfrac{23}{299} $=$ \dfrac{1}{13} $.
Therefore, A and B together can complete the work in 13 days.