What is motion?
- It is defined as change of position with time. Example:- the moving blades of a fan, the moving car.
- motion is a relative term.
Motion of an object is always with respect to some frame if reference.
Rectilinear motion:- Motion along a straight line
Circular motion:- Motion along a circular path.
Describing motion of an object?
In order to describe the motion, position needs to be defined.
Position:- Point at which the object is located at a particular instant of time.
Path length and displacement:-
Path length :- Total distance an object travels from initial to final position .
Displacement :- Shortest distance between initial and final position.
Path length can be equal to displacement or greater than displacement.
Path length cannot be zero whereas displacement can be zero.
Scalar and vector quantity:-
Scalar quantity:- quantity with magnitude alone and no direction.
Example:- path length, speed .
Magnitude:- numerical value that defines the measure of a quantity.
Vector quantity:- quaniqua with both magnitude and direction.
Example:- displacement, velocity .
Speed and velocity:-
Speed :-
- rate at which an object covers distance.
- Scalar quantity
- Path length /time taken = speed
- SI unit is m/sec
Velocity:-
- Rate at which an object changes its position.
- Speed with direction
- Vector quantity
- Displacement /time taken= velocity
- SI unit is m/sec.
Distance - Time graphs :-
Slope = speed of object
Uniform and non uniform motion:-
Uniform motion:- A body is said to be in uniform motion, if it covers equal distance in equal intervals of times.
Non uniform motion:- A body is said to be in non uniform motion if it covers unequal distances in equal intervals of time.
Interpretation from different- Distance- time graphs:-
1. Object at rest i.e no motion occur.
2. Object covers equal distance in equal interval of time i.e uniform motion
3. Object is at rest for first one second .after that it shows uniform motion.
4 Non uniform motion.
Average speed:-
- Measure of total distance covered in a given time.
- In uniform motion , speed is equal to average speed.
Average speed and instantaneous speed:-
Average speed:- Measure of total distance covered in a given time.
Instantaneous speed:- Measure of distance covered at every instant of time.
Average velocity:- Measure of total displacement covered in a given time.
It can be zero
Instantaneous velocity:-Measure of displacement covered at every instant of time.
Acceleration:-
- Rate of change of velocity .
- Vector quantity
- SI unit is m/sec^2
- It can be zero .negative and positive.
Slope of velocity -Time graph :- Acceleration
Acceleration = zero
When 1. Object is at rest ,2. Object is in uniform motion.
Graphical representation of motion:-
Interpretation from different velocity -Time graph:-
Area under V-T curve:- Displacement:-
Conclusion from PT graphs and VT graphs:-
- Slope of P-T graph-> velocity
- Slope of V-T graph-> Acceleration
- Area under V-T graph- for uniformly accelerated motion = displacement.
Equations of motion:- These are the sets of equations which are describe motion of object .
These establish relation between:-
- Displacement ,s
- Time taken,t
- Initial velocity,u
- Final velocity,v
- Acceleration,a
Three kinematics equations are:-
- v= u+ at
- s= ut +1/2at^2
- v^2 = u2 + 2as
Derivation:- v= u+at
- Let initial velocity= u
- Final velocity. = v
- Time taken t
a = v-u/t
On cross multiple
at= v-u
at + u = v
Or v= u+at
Second equation of motion:-
s = ut + 1/2 at^2
s= area under the graph
s= area of rectangle + area of triangle
s= length * breadth + 1/2 base * height
s = ut + 1/2 t(v-u)
s = ut + 1/2 t(at) ....because at = v-u
s = ut + 1/2 at^2
Third equation.of motion:- v^2 = u^2 + 2as
s = area under the graph
s = area of trapezium
s = 1/2 [sum of parallel side] * height
s = 1/2 [u + v] *t
s = 1/2 [v + u] [v-u] /a
s= [v^2 - u^2]/2a
On cross multiple
2as = v^2 -u^2
2as + u^2 = v^2
Or
v^2 = u^2 + 2as
Uniform circular motion:-
A body is said to be in uniform circular motion if it moves in a circular path with uniform acceleration.
Numerical:-
1. An athlete covers one round of a circular track of diameter 200 m in 40 second. What will be the distance covered and the displacement at the end of 2 min 20 second?
Solution:-
Diameter = 200m
Time taken to cover 1 round= 40 sec
So distance covered in 40 sec = 2πr = 2*22/7*100= 4400/7
Distance covered in 140 sec = (4400*140)/7*40= 2200 m ° 2.2km
(ii) Rounds in 40 sec = 1
Rounded in 140 sec = 1*140/40 = 7/2= 3.5
So displacement= 0.5 round = diameter = 200m
2.Usha swims in a 90m long pool . She covers 180m
Solution:-
Average speed= total distance/total time = 180/60 =3m/sec
Average velocity= total displacement/total time = 0/60 = zero
3. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging.
(a) from A to B and
(b) from A to C
Solution :-
From A to B
Average speed = 300/(120+50) = 300/170=30/17 m/sec
Average velocity = 30/17m/sec
b) from A to C
Average speed= 300+100/(170+60)=400/230=40/23 m/sec = 1.73 m/sec
Average velocity= 200/230=20/23=0.87m/sec
4. The odometer of a car reads 200 km at the start of a trip took 8hours . Calculate the average speed of car in Km/Hr and m/s
Solution:-
Distance covered = 2400-2000=400km
Time taken= 8hours
Average speed= distance /time = 400/8= 50km/Hr= 50*1000/3600= 13.9m/s
5. Abdul while driving to school , computes the average speed for his trip to be 20km/hr . On his return trip along the same route, there is less traffic and the average speed is 40km/he . What is the average speed for Abdul's trip.
Solution:-
Average speed from home to school = 20km/hr
Average spree from school to home= 40km/hr
Let the distance from school to home or vice versa = dkm
time taken from home to school ,t1= distance/speed=(d/20)hour
time taken from school to home,t2= distance/speed=(d/40)hr
Total time= d/20+d/40=3d/40
Average speed=total distance/total time
=(d +d)*40/3d=2d*40/3d=80/3km/hr
6. A bus decrease it's speed from 80 Km/he to 60km/Hr in 5 second. Find acceleration of the bus.
u=80km/hr = 80*5/18 =200/9 m/sec
v= 60km/hr = 60*5/18= 30*5/9=150/9m/sec
t=5sec
a=v-u/t=(150/9-200/9)/5=-50/9*5=-10/9=-1.12m/s^2
7. A train starting from a railway station and moving with uniform acceleration attains a speed of 40km/hr in 10 minutes. Find its acceleration.
Solution:-
u = 0
v = 40km/hr =40*1000/3600=400/36=100/9m/s
t=10min =10*60= 600sec
a= (v-u) /t = (100/9-0)/600 = 100/(9*600)=1/54m/s^2= 0.017m/s^2
8. Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6m/s in 30 second. Then he applied brakes such that the velocity of bicycle comes down to 4m/s in next 5 second. Calculate the acceleration of bicycle in both cases.
Solution:-
u =0
v = 6m/s
t= 30sec
a = (v-u)/t = (6-0)/30=6/30=1/5 m/s^2
Now u = 6m/s
v' =4m/s
t=5sec
a= (v-u)/t
a=(4-6)/5 = -2/5 m/s^2
9. The speed time graph for a car is shown in fig :-
(a) find how far does the car travel in the first 4 seconds.
Shade the area on the graph that represents the distance travelled by the car during the period
(b) which part of the graph represents uniform motion of the car?
10. A bus starting from rest moves with a uniform acceleration of 0.1 m/s^2 for 2 minutes.
Find (a) speed acquired and (b) distance travelled
11. A train is travelled at a speed of 90km/Hr . Breaks are applied so as to produce a uniform acceleration of -0.5 m/s^2 . Find how far the train will go before it is brought to rest.
Solution:-
12. A trolley while going down an inclined plane has an acceleration of 2 cm/s^2 . What will be it's velocity 3s after the start.
Solution :-
a= 2cm/s^2 , u=0 , v=? ,t=3sec
v = u + at
v = 0 +2(3)
v = 6cm/sec
13. A racing car has a uniform acceleration of 4 m/s ^2 . What distance will it cover in 10 sec after start?
Solution:-
a= 4m/s^2 , s=? , u= 0, t= 10sec
s = ut + 1(at^2)/t
s = 0 + 1(4*10*10)/2
s = 200m
14. A stone is thrown in a vertically upward direction with a velocity of 5m/s . If the acceleration of the stone during its motion is 10m/s^2 in downward direction, what will be the height attained by stone and how much time will it take to reach there?
Solution:-
u=5m/s , a=-10m/s^2 , v=0 (reaches maximum height)
Let maximum height = s
v= u +at
0= 5 - 10t
5/10 =t
0.5 sec =t
Now
v^2 = u^2 +2as
0 = 5^2 + 2(-10)s
0=25 -20s
20s=25
s=5/4 =1.25m
15. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s^2 for 8 sec . How far does boat travel during the time?
Solution:-
u = 0 , a= 3m/s^2 , t = 8sec. , s=?
s = ut + 1(at^2)/2
s = 0 + 1(3*8*8)/2
s = 0 + 96
s = 96m .
16. A ball is gently dropped from a height of 20 m . If it's velocity increase uniformly at the rate of 10 m/s^2 , with what velocity will it strikes the ground? After what time will it strikes the ground?
Solution:-
s = 20m , a = 10m/s^2 , u = 0 , v=?
v^2 = u^2 + 2as
v^2 = 0 + 2(10*20)
v^2 = 400
v = 20m/s
Now
v = u + at
20 = 0 + 10(t)
20/10 = t
2 sec= t
17. A train starting from rest attains a velocity of 72km/hr in 5mins . Assuming that the acceleration is uniform, find (I) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
u=0
v=72km/hr =72*5/18=4*5=20 m/sec
t=5min=5*60=300sec
a=v-u/t =20-0/300=20/300=2/30=1/15m/s^2
(ii) v2-u2=2as
400-0=2*(1/15)s
(400*15)/2=s
3000m=s
3km=s
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