1.
Time and Distance
Introduction:
This
chapter mainly deals with problem related among Distance, Time and Speed of the
objects like body, birds, trains, persons, motor vehicles and submarines like
any moving bodies.
Basic Formulas:
1. Speed=Distance/Time ;
Speed(s), Distance(d), Time(t)
2. s∞1/t => s1t1=s2t2 =>
s1/s2=t2/t1
3. If speeds are constant then d1/t1=d2/t2
=> d1/d2=t1/t2
4.
If reaching times are same
d/s=constant d1/s1=d2/s2 =>
s1/s2=d1/d2
5. when a motor car covers
a distance at s1 km/hr and an equal distance at s2 km/hr then the average speed for the
total journey is S=(2s1s2)/s1+s2
Conversion Factor:
1. from km/hr to m/sec:
When a motor car running with speed x
km/hr its speed in m/sec is (x*5/18)m/sec
Ex:
54km/hr 54*5/18=15m/sec
2. from m/sec to km/hr:
When a car
is running with speed y m/sec then its speed in km/hr is y*18/5 km/hr
Ex: car speed 50m/sec then 50*18/5=180
km/hr.
Conversion of Time:
1Hr=60min=3600sec,
1day=24hr=24*60*60sec=86400sec
Ex:
1Hr.40min => 1hr*40*1/60hr=5/3hr.
2min.30sec
=> 2*30/60=1min
Note: when the
speeds are in the same direction the resultant speed is S=s1-s2
When the
speeds are in the opposite direction the resultant speed is S=s1+s2
PROBLEMS:
1. While
covering a distance of 240km, a car driver noticed that after driving for 1hr
and 40min, the distance covered by him was 5km of the remaining distance. What
was the speed of the car?
Sol. Let x be the speed of the car in km/hr
The distance
traveled in 1hr and 40min = xkm/hr*1hr.40min*1/60hr
= 5/3x
km
In the same time we found the distance covered
is=5/7(240-5x/3) km
Distances
covered are same, therefore
5x/3=5/7(240-5x/3)
7x=3(240-5x/3) = (3*240-(5x*3)/3)
7x=720-5x
7x+5x=720
12x=720
X=720/12=60km/hr
X=60km/hr.
2. Two men starting from the same place walk at the rate of
10km/hr and 8km/hr respectively. Find the separation between them at the end of
7/2 hr if a) They walk in the opposite
direction b) they walk in the same
direction
Sol. A) When they walk in opp. Direction the speeds are
added S=s1+s2=10+8=18km/hr
The separation
b/w them is 18km in one hr.
In 7/2hr they are
apart, 18*7/2=63km
B) When they walk in
same direction speeds are subtracted S=s1-s2=10-8=2km/hr
They are
separated in 2km,
In 7/2 hr they
are apart=2*7/2=7km.
Solve the Problems:
- A
train traveling 25km/hr leaves
a) 430km b) 437.5km c) 500km d) 117km
- A is faster than B. A and B walk 24km the sum of their speed is 7km/hr and the sum of times taken by them is 14hr then A’s speed is
a) 3km/hr b) 4km/hr c) 5km/hr d) 7km/hr
- In covering a distance of 30km, shaker takes 2hr more than suman. If shaker doubles his speed then he would take 1hr less than suman.Then shaker’s speed is.
a) 5kmph
b) 6kmph c) 6.25kmph d) 7.5kmph
- The jogging in a sports complex is 726m in circumference. Sachin and his wife starts from the same point and walk in opposite direction at 4.5km/hr and 3.75km/hr.they will meet first time in
a) 4.9min b) 5.28min c) 5.5min d) 6min
5. A can complete a journey in
10hrs. He travels first half of the journey at the rate of 21km/hr and second
half at the rate of 24km/hr. The total journey in km
a) 220km b) 224km c) 230km d) 234km
- A person travels from P to Q at a speed of 40kmph and returns by increasing his speed by 50%. What is his average speed for both the trips?
a) 36kmph b) 45kmph c) 48kmph d) 50kmph
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