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An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Yes, this is very much possible.
An object can have zero displacement even when it has moved through a distance. This happens when final position of the object coincides with its initial position.
a) A person moves around park and stands on place from where he started then here displacement will be zero but distance travelled will not be zero
b) If you go around a circle and come back to the starting point, then displacement will be zero but distance will not be
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Let ABCD is the square field and Farmer starts at point A and moves in the anticlockwise direction
Given in the problem
Side of the square field= 10m
Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40s.
So speed of the farmer= distance travelled/ Time taken = 40 /40 = 1m/s
Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?
Distance travelled in 140s =speed × time= 1 × 140 m = 140 m
Now, number of Lap to cover 140 along the boundary= Total Distance / Perimeter
= 140 m / 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.
So displacement = Magnitude of AC = Diagonal of the square of side 10 m
Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
None of the statement is true for displacement as
1) displacement can be zero.
2) displacement is less than or equal to the distance travelled by the object.
Distinguish between speed and velocity.
Speed is the distance travelled by an object in a given interval of time.
Velocity is the displacement of an object in a given interval of time.
Speed = distance / time
Velocity = displacement / time
Speed is scalar quantity i.e. it has only magnitude.
Velocity is vector quantity i.e. it has both magnitude as well as direction.
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
The magnitude of average velocity of an object is equal to its average speed, only when an object is moving in a straight line.
What does the odometer of an automobile measure?
The distance covered by an automobile.
What does the path of an object look like when it is in uniform motion?
straight line path.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108
Speed= 3 × 108
Time= 5 min = 5 x 60 = 300 secs.
We know that
Distance= Speed x Time
Distance= 3 × 108
x 300 secs. = 9 x 1010
When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
A body is said to be in non-uniform acceleration if the rate of change of its velocity is not constant.
A bus decreases its speed from 80 km h−1
to 60 km h−1
in 5 s. Find the acceleration of the bus.
We have to very careful with units while doing the numerical questions.
Here we would convert km/h into m/s
Initial Speed (u) = 80 km/h= (80 ×1000)/3600 m/s = 22.22 m/s
Final Speed (u) = 60 km/h= (60 ×1000)/3600 m/s = 16.66 m/s
Acceleration is given by
= Change in speed/ time taken
Here negative sign denotes de-acceleration
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1
in 10 minutes. Find its acceleration.
Initial velocity of the train (u) =0
Final velocity of the train (v)=40 km/h = 11.11 m/s
Time taken(t) =10 min= 600s
Acceleration is given by
= Change in speed/ time taken
= .0185 m/s2
What is the nature of the distance - 'time graphs for uniform and non-uniform motion of an object?
When the motion is uniform, the distance time graph is a straight line with a slope.
When the motion is non uniform, the distance time graph is not a straight line. It can be any curve.
What can you say about the motion of an object whose distance - time graph is a straight line parallel to the time axis?
It is at rest
What can you say about the motion of an object if its speed - 'time graph is a straight line parallel to the time axis?
It means speed is constant and object is moving uniformly
What is the quantity which is measured by the area occupied below the velocity -time graph?
The area below velocity-time graph gives the distance covered by the object.
A bus starting from rest moves with a uniform acceleration of 0.1 m /s2
for 2 minutes. Find
(a) the speed acquired
(b) the distance travelled.
Initial speed of the bus(u) = 0
Acceleration(a) = 0.1 m/s2
Time taken(t) = 2 minutes = 120 s
According the first equation of motion
v= u + at
v= 0 + 0.1 × 120
v= 12 m/s
According to the third equation of motion:
Where, s is the distance covered by the bus
= 2(0.1) s
s = 720 m
A train is travelling at a speed of 90 km h−1
. Brakes 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.
Initial speed of the train(u)= 90 km/h = 25 m/s
Final speed of the train(v) = 0 (As finally the train comes to rest)
Acceleration = - 0.5 m/s2
According to third equation of motion:
+ 2 (- 0.5) s
Where, s is the distance covered by the train
s = 625/2(.5) = 625m
The train will cover a distance of 625 m before it comes to rest.
A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2
. What will be its velocity 3 s after the start?
Initial Velocity of trolley(u)= 0 cm/s
Acceleration(a)= 2 cm /s2
Time, t= 3 s
We know that final velocity is given by Ist equation of motion
v= u + at
v=0 + 2 x 3 cm/s
Therefore, The velocity of train after 3 seconds = 6 cm/s
A racing car has a uniform acceleration of 4 m/s2
. What distance will it cover in 10 s after start?
Initial Velocity of the car(u)=0 m/s
Acceleration(a)= 4 m /s2
Time(t)= 10 s
We know Distance is given by 2nd
equation of motion
s= ut + (1/2) at2
Therefore, Distance covered by car in 10 second
s= 0 × 10 + (1/2) × 4 × (10)2
= 200 m
A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m /s2
in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Initial velocity of stone(u)=5 m /s
Acceleration(a)= -10 m /s2
(as acceleration is opposite to direction of motion)
Final velocity of stone (v) =0 m/s
We know as per 3rd
equation of uniform motion
2(-10) s= 0- (5)2
s=25/20= 12.5 m
We know as per Ist equation of uniform motion
V=u + at
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Given Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 x (22 / 7) x 100 = 4400/7 m
Speed of the athlete (s) = Distance / Time
= (4400) / (7 x 40) = 110/7 m/s
Now Distance covered in 140 s
= Speed × Time
= (110/7) x (140)
= 2200 m
Number of round in 40 s =1 round
Number of round in 140 s =140/40
Let us assume athlete start at point A on the circular ground
After taking start from position A, the athlete will be at position B after 3 1
Hence, Displacement of the athlete with respect to initial position at B= AB
= Diameter of circular track
= 200 m
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging?
(a) from A to B
(b) from A to C?
Total Distance covered from AB = 300 m
Total time taken = 2 x 60 + 30 s
a) Average Speed from AB = Total Distance / Total Time
=300 / 150 m/s
=2 m /s
Average Velocity from AB =Displacement AB / Time = 300 / 150 m s-1
=2 m s-1
b) Total Distance covered from AC =AB + BC
=300 + 100 m
Total time taken from A to C = Time taken for AB + Time taken for BC
= 210 s
Average Speed from AC = Total Distance /Total Time
= 400 /210 m /s
= 1.904 m /s
Now Displacement (S) from A to C = AB - BC
= 300-100 m
= 200 m
Time (t) taken for displacement from AC = 210 s
Average Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m /s
= 0.952 m /s
Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1
. On his return trip along the same route, there is less traffic and the average speed is 30 km h−1
. What is the average speed for Abdul’s trip?
let the distance Abdul commutes while driving from Home to School = d
Let us assume time taken by Abdul to commutes this distance = t1
Now Distance Abdul commutes while driving from School to Home = d
Let us assume time taken by Abdul to commutes this distance = t2
Average speed from home to school v1
= 20 km/h
Average speed from school to home v2
= 30 km/h
we know Time taken from Home to School t1
Similarly Time taken from School to Home t2
Total distance from home to school and backward = 2d
Total time taken from home to school and backward (T) = d/20+ d/30
Average speed (Vav
) for covering total distance (2S) = Total Distance/Total Time
= 2d / (d/20 +d/60)
= 2d / [(2d+d)/60]
= 120d / 5d
= 24 km/h
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m /s2
for 8.0 s. How far does the boat travel during this time?
Initial velocity of motor boat(u) = 0
Acceleration of motor boat, a = 3.0 m s-2
Time under consideration, t = 8.0 s
We know that Distance travelled as per 3rd
law of uniform motion
s = ut + (1/2)at2
The distance travel by motorboat
s= 0 x 8 + (1/2)3.0 x (8)2
= (1/2) x 3 x 8 x 8 m
= 96 m
A driver of a car travelling at 52 km h−1
applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1
in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
First driver car Initial speed (uA
) = 52km/h = 52 x 1000/3600 = 14.4 m/s
Final Speed (vA
) = 0 m/s
Time taken (tA
) = 5s
Second driver car Initial speed (uB
) = 3km/h = 3 x 1000/3600 = 0.83 m/s
Final Speed (vB
) = 0 m/s
Time taken (tB
) = 10s
Plotting on Speed-time graph, the graph PR and SQ are the Speed-time graph for given two cars with initial speeds 14.4 m/s and .83 m/s respectively.
Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 14.4 m/s
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x .83 m/s= 4.16 m
Below Figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d)How far has B travelled by the time it passes C?
Important points to note
Speed is given by = Distance /time = Y-axis/X axis = Slope
So the object with steepest slope will have fastest speed
(a) Object B Since slope of object B is greater than objects A and C, it is travelling the fastest.
(b) No as all the three lines never intersect at the same point
c) When B and A meets at point, the Object C is at a point, which is approximately 8km on distance axis.
(d) B meets C at point . The position on distance-axis is 9 units away from origin O. One unit on distance-axis = 4/7 (7 units = 4km).
So Object B is = 9 x 4/7 = 5.14 km.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2
, with what velocity will it strike the ground? After what time will it strike the ground?
Initial Velocity of ball(u) =0
Distance or height of fall(s) =20 m
Downward acceleration, a =10 m /s2
As we know from 3rd
equation of uniform motion
= 2as+ u2
Final velocity of ball, v = 20 m/s
As we know from 1st
equation of uniform motion
t = (v-u)/a
Time taken by the ball to strike = (20-0)/10
= 2 seconds
The speed-time graph for a car is shown is given below
(a) Find out how far the car travels 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?
The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled by the car in the first 4 s.
The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.
State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2
When a car is moving in a circular track, its acceleration is perpendicular to its direction.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Radius of the circular orbit, r= 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v= (2π r)/t
=3073.74 m /s
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