ad

$\text{speed} = \dfrac{\text{distance}}{\text{time}}$

$\text{distance} = \text{speed}×\text{time}$

$\text{time} = \dfrac{\text{distance}}{\text{speed}}$

2. Convert kilometres per hour(km/hr) to metres per second(m/s)
$\text{distance} = \text{speed}×\text{time}$

$\text{time} = \dfrac{\text{distance}}{\text{speed}}$

$x\text{ km/hr} = x×\dfrac{5}{18} \text{m/s}$

3. Convert metres per second(m/s) to kilometres per hour(km/hr)
$x \text{ m/s} = x×\dfrac{18}{5} \text{ km/hr}$

4. 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$= \dfrac{2xy}{x+y}$ kmph

5. Relation Between Distance, Speed and Time (5.1) Speed and time are inversely proportional (when distance is constant)

$\implies \text{speed} \propto \dfrac{1}{\text{time}}$ (when distance is constant)

(5.2) 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

$\dfrac{1}{a}:\dfrac{1}{b}=b:a$

(5.3) Assume two objects A and B start at the same time in opposite directions from P and Q respectively. After passing each other, A reaches Q in $a$ seconds and B reaches P in $b$ seconds. Then,

Speed of A : Speed of B $=\sqrt{b}:\sqrt{a}$

(5.4) An object covered a certain distance at a speed of $v$ kmph. If it had moved $v_1$ kmph faster, it would have taken $t_1$ hours less. If it had moved $v_2$ kmph slower, it would have taken $t_2$ hours more. Then,

$v=\dfrac{v_1v_2(t_1+t_2)}{v_1t_2-v_2t_1}\text{ kmph}\\ x=vt_1\left(1+\dfrac{v}{v_1}\right)\text{ km}$

Special Case:

If $t_1=t_2,~~v=\dfrac{2v_1v_2}{v_1-v_2}\text{ kmph}$ 6. Relative Speed (6.1) If two objects are moving in the same direction at $v_1$ m/s and $v_2$ m/s respectively where $v_1 \gt v2,$ then their relative speed $=(v_1-v_2)$ m/s

(6.2) Consider two objects A and B separated by a distance of $d$ metre. Suppose A and B start moving in the same direction at the same time such that A moves towards B at a speed of $a$ metre/second and B moves away from A at a speed of $b$ metre/second where $a \gt b.$ Then,

relative speed $=(a-b)$ metre/second

time needed for A to meet B $=\dfrac{d}{a-b}$ seconds

(6.3) If two objects are moving in opposite directions at $v_1$ m/s and $v_2$ m/s respectively, then their relative speed $=(v_1+v_2)$ m/s

(6.4) Consider two objects A and B separated by a distance of $d$ metre. Suppose A and B start moving towards each other at the same time at $a$ metre/second and $b$ metre/second respectively. Then,

relative speed $=(a+b)$ metre/second

time needed for A and B to meet each other $=\dfrac{d}{a+b}$ seconds

unknown

2014-07-29 15:55:54

A hare and tortoise have a race along a 100 yards diameter. The tortoise goes in one direction and the hare in the other.Hare starts after tortoise has covered 1/4 of the distance.they meet when hare has covered only 1/5 of the distance. By what factor should hare increase his speed so as to tie the race?

Neeraj Jain

2014-07-29 22:44:50

hare starts only after tortoise has covered 1/4 of the distance and they meet when hare has covered only 1/5 of the distance.

Total distance travelled by the hare = 1/5

Total distance travelled by the tortoise = 1 - 1/5 - 1/4 = 11/20

distance Traveled hare / distance travelled by tortoise = (1/5) / (11/20) = 4/11

Since distance = speed * time and both take same distance to travel this distance,

**speed of hare/ speed of tortoise = 4/11 --------------(1)**

Remaining distance to be travelled by the hare = 1 - 1/5 = 4/5

Remaining distance to be travelled by the tortoise = 1/5

distance to be travelled by the hare / distance to be travelled by the tortoise = (4/5)/(1/5) = 4

Since both must travel these distances in same time,

**new speed of hare /speed of the tortoise = 4 --------------(2)**

Divide equations 2 by equation 1

**new speed of the hare /****speed of hare = 4/(4/11) = 11**

Required factor is 11

Total distance travelled by the hare = 1/5

Total distance travelled by the tortoise = 1 - 1/5 - 1/4 = 11/20

distance Traveled hare / distance travelled by tortoise = (1/5) / (11/20) = 4/11

Since distance = speed * time and both take same distance to travel this distance,

Remaining distance to be travelled by the hare = 1 - 1/5 = 4/5

Remaining distance to be travelled by the tortoise = 1/5

distance to be travelled by the hare / distance to be travelled by the tortoise = (4/5)/(1/5) = 4

Since both must travel these distances in same time,

Divide equations 2 by equation 1

Required factor is 11

0
0

srinivas

2014-05-12 05:31:51

two cyclistas start from the same place in opposite directions. one goes towards nouth at 18kmph and the other goes towards south at 20kmph. what time will they take to be 47.5km apart..

Anu Lakshmi

2014-05-15 20:26:41

relative speed = 18+20 = 38 kmph

distance = 47.5 km

time taken = distance/speed = 47.5 / 38 hr = 1.25 hr = 75 minutes

distance = 47.5 km

time taken = distance/speed = 47.5 / 38 hr = 1.25 hr = 75 minutes

0
0

Nimisha

2013-08-09 17:27:35

Simple!

The ratio of the speeds of A and B is 1 : 2. What is the ratio of the time taken by them to cover 200 km?

As per the formula, you can directly get answer as 1/1 : 1/2 = 2 : 1

__I will also explain how the result is derived__

Ratio of the speeds of A and B is 1 : 2

hence, we can assume that the speed of A = k and speed of B = 2k

Time taken by A to cover 200 km = distance/speed = 200/k

Time taken by B to cover 200 km = distance/speed = 200/2k

Required ratio = 200/k : 200/2k = 1 : 1/2 = 2 : 1

Does it help?

The ratio of the speeds of A and B is 1 : 2. What is the ratio of the time taken by them to cover 200 km?

As per the formula, you can directly get answer as 1/1 : 1/2 = 2 : 1

Ratio of the speeds of A and B is 1 : 2

hence, we can assume that the speed of A = k and speed of B = 2k

Time taken by A to cover 200 km = distance/speed = 200/k

Time taken by B to cover 200 km = distance/speed = 200/2k

Required ratio = 200/k : 200/2k = 1 : 1/2 = 2 : 1

Does it help?

0
0

preview