Page No:42
Question 38:
Show by means of graphical method that: v = u + at, where the symbols have their usual meanings.
Solution :
Consider the velocitytime graph of a body shown in figure.
The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.
Now, Initial velocity of the body, u= OA —–(1)
And,Final velocity of the body, v=BC ——(2)
But from the graph BC =BD + DC
Therefore, v=BD + DC ——(3)
Again DC = OA
So,v =BD + OA
Now, from equation (1), OA =u
So, v=BD + u ——(4)
We should find out the value of BD now. We know the slope of a velocitytime graph is equal to the acceleration, a.
Thus,Acceleration, a= slope of line AB
_{or a = BD/AD}
But AD =OC= t, so putting t in place of AD in the above relation, we get:
or BD=at
Now, putting this value of BD in equation(4), we get:
v= u+ at
consider the velocity  time graph of a body shown in figure:
the body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.
Now initial velocity of the body, u = OA  (1)
and, final velocity of the body, v = BC  (2)
But from the graph BC = BD+DC
Therefore v = BD+DC (3)
Again DC = OA
So, v = BD+ OA
Now, from equation (1), OA = u
So, v= BD+ u (4)
we should find out the value of BD now. We know the slope of the velocity time graph is equal to the acceleration , a
Thus acceleration a = slope of line AB
or a = BD/AD
But AD = OC= t, so putting t in place of AD in the above relation, we get:
or BD = at
Now, Putting the value of BD in equation (4), we get
v= u+at
Question 39:
Show by using the graphical method that: , where the symbols have their usual meanings.
Solution :
Consider the velocitytime graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC.
Suppose the body travels a distance s in time t. In the figure, the distance travelled by the body is given by the area of the space between the velocitytime graph AB and the time axis OC, which is equal to the area of the figure OABC. Thus:
Distance travelled = Area of figure OABC
= Area of rectangle OADC + area of triangle ABD
Now, we will find out the area of rectangle OADC and area of triangle ABD.
(i) Area of rectangle OADC =OA x OC
= u x t
=ut
(ii) Area of triangle ABD= _{(1/2)}x Area of rectangle AEBD
= _{(1/2) }x AD x BD
= _{(1/2)} x t x at
= _{(1/2)} at^{2}
Distance travelled, s = Area of rectangle OADC + area of triangle ABD
Consider the velocitytime graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity become v which is equal to BC in the graph. The time t is represented by OC.
Suppose the body travel a distance in time t . in the figure, the distance travelled by the body is given by the area of the space between the velocitytime graph AB and the time axis OC, which is equal to the area of the figure OABC. Thus:
Distance travelled = Area of figure OABC
= Area of rectangle OADC+ area of triangle ABD
Question 40:
Derive the following equation of motion by the graphical method : v^{2} = u^{2} + 2as, where the symbols have their usual meanings.
Solution :
Consider the velocitytime graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.
The distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.
Distance travelled, s= Area of trapezium OABC
Now, OA + CB = u + v and OC =t Putting these values in the above relation, we get:
Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
Consider the velocity of time graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. the time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE
the distance travelled s by a body in time t is given by the area of the figure OABC which is trapezium.
Now, OA +CB = u+v and OC = t putting these values in the above relation, we get:
Eliminate t from the above equation. This can be done obtaining the value of t from the first equation of motion.
thus v = u +at (first equation of motion)
and at = v
Page No:43
Question 53:
The graph given alongside shows the positions of a body at different times. Calculate the speed of the body as it moves from :
(i) A to B,
(ii) B to C, and
(iii) C to D.
Solution :
(i) The distance covered from A to B is( 30) =3 cm
Time taken to cover the distance from A to B =(5 2) =3s
(ii) The speed of the body as it moves from B to C is zero.
(iii) The distance covered from C to D is (73)=4 cm
Time taken to cover the distance from C to D = (97)=2s
Question 54:
What can you say about the motion of a body if:
(a) its displacementtime graph is a straight line ?
(b) its velocitytime graph is a straight line ?
Solution :
(a) The body has a uniform velocity if its displacementtime graph is a straight line.
(b) The body has a uniform acceleration if its velocitytime graph is a straight line.
Question 55:
A body with an initial velocity x moves with a uniform acceleration y. Plot its velocitytime graph.
Solution :
Question 56:
Given alongside is the velocitytime graph for a moving body :
Find :
(i) Velocity of the body at point C.
(ii) Acceleration acting on the body between A and
(iii) Acceleration acting on the body between B and C.
Solution :
(i) BC represents uniform velocity. From graph, we see that the velocity of the body at point C = 40km/h
(ii) Acceleration between A and B = slope of line AB
(iii) BC represents uniform velocity, so acceleration acting on the body between B and C is zero.
Question 57:
A body is moving uniformly in a straight line with a velocity of 5 m/s. Find graphically the distance covered by it in 5 seconds.
Solution :
Distance travelled = Area of rectangle OABC
= OA x OC
= 5 x 5 =25 m
Question 58:
The speedtime graph of an ascending passenger lift is given alongside.
What is the acceleration of the lift:
(i) during the first two seconds ?
(ii) between second and tenth second ?
(iii) during the last two seconds ?
Solution:
(i) Acceleration during first two seconds
(ii) Acceleration between second and tenth second is zero, since the velocity is constant during this time.
(iii) Acceleration during last two seconds
Question 59:
A car is moving on a straight road with uniform acceleration. The speed of the car varies with time as follows :
Draw the speedtime graph by choosing a convenient scale. From this graph :
(i) Calculate the acceleration of the car.
(ii) Calculate the distance travelled by the car in 10 seconds.
Solution :
(i) Acceleration of the car = slope of line AB
(ii) Distance travelled by the car in 10 s = area of trapezium OABC
Question 60:
The graph given alongside shows how the speed of a car changes with time:
(i) What is the initial speed of the car ?
(ii) What is the maximum speed attained by the car ?
(iii) Which part of the graph shows zero acceleration ?
(iv) Which part of the graph shows varying retardation ?
(v) Find the distance travelled in first 8 hours.
Solution :
(i) Initial speed of the car=10km/h
(ii) Maximum speed attained by the car= 35km/h
(iii) BC represents zero acceleration.
(iv) CD represents varying retardation.
(v)
Distance travelled in first 8 hrs:
s= Area of trapezium OABF + Area of rectangle BCEF
Question 61:
Three speedtime graphs are given below :
Which graph represents the case of:
(i) a cricket ball thrown vertically upwards and returning to the hands of the thrower ?
(ii) a trolley decelerating to a constant speed and then accelerating uniformly ?
Solution :
(i) Graph (c): The speedof the ball goes on decreasing uniformly as it moves upward, reaches zero at the highest point, and then increases uniformly as it moves downward.
(ii) Grap(a): The speed of the trolley decreases uniformly, then it moves at a constant speed, and then the speed increases uniformly.
Page No:44
Question 62:
Study the speedtime graph of a car given alongside and answer the following questions:
(i) What type of motion is represented by OA ?
(ii) What type of motion is represented by AB ?
(iii) What type of motion is represented by BC ?
(iv) What is the acceleration of car from O to A ?
(v)What is the acceleration of car from A to B ?
(vi) What is the retardation of car from B to C ?
Solution :
(i) OA represents uniform acceleration
(ii) AB represents constant speed.
(iii) BC represents uniform retardation.
(iv) Acceleration of car from O to A = slope of line OA
(v) Acceleration of car from A to B is zero as it has uniform speed during this time.
(vi) Retardation of car from B to C = slope of line BC
Question 63:
What type of motion is represented by each one of the following graphs ?
Solution :
(i) Graph (a) represents uniform acceleration.
(ii) Graph (b) represents constant speed.
(iii) Graph (c) represents uniform retardation.
(iv) Graph (d) represents nonuniform retardation.
Question 64:
A car is travelling along the road at 8 ms^{1}. It accelerates at 1 ms^{2} for a distance of 18 m. How fast is it then travelling ?
Solution :
Initial velocity, u=8m/s
Acceleration, a=1m/s^{2}
Distance, s=18m
Question 65:
A car is travelling at 20 m/s along a road. A child runs out into the road 50 m ahead and the car driver steps on the brake pedal. What must the car’s deceleration be if the car is to stop just before it reaches the child ?
Solution :
Initial velocity, u=20m/s
Final velocity, v=0m/s
Distance, s=50m
66 videos355 docs97 tests

1. What is motion in physics? 
2. What are the different types of motion? 
3. How is speed different from velocity? 
4. What is the difference between distance and displacement? 
5. How does acceleration affect motion? 
66 videos355 docs97 tests


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