Chapter 3 Motion in a Straight Line Solutions
Question - 11 : - Read each statement below carefully and statewith reasons and examples, if it is
true or false;
A particle in one-dimensional motion
(a) with zero speed at an instant mayhave non-zero acceleration at that instant
(b) with zero speed may have non-zerovelocity,
(c) with constant speed must have zeroacceleration,
(d) with positive value of accelerationmust be speeding up
Answer - 11 : -
(a) True
(b) False
(c) True (if theparticle rebounds instantly with the same speed, it implies infiniteacceleration which is unphysical)
(d) False (true onlywhen the chosen positive direction is along the direction of motion)
Question - 12 : - A ball is dropped from a height of 90 m on a floor. At each collisionwith the floor, the ball loses one-tenth of its speed. Plot the speed-timegraph of its motion between t = 0 to 12 s.
Answer - 12 : -
Height from which theball is dropped = 90 m
The initial velocityof the ball, u = 0
Let v be the finalvelocity of the ball
Using the equation
v2 – u2 =2as ——–(1)
v12 –0 = 2 x 10 x 90
v1= 42.43m/s
Time taken for firstcollision can be given by the equation
v = u + at
42.43 = 0 + (10) t
t1 =4.24 s
The ball lossesone-tenth of the velocity at collision. So, the rebound velocity of the ball is
v2= v –(1/10)v
v2 =(9/10) v
v2= (9/10)(42.43)
= 38.19 m/s
Time taken to reachmaximum height after the first collision is
v = u + at
38.19 = 0 + (10)t2
t2 =3.819 s
Total time taken bythe ball to reach the maximum height is
T = t1 +t2
T = 4.24+ 3.819 = 8.05 s
Now the ball willtravel back to the ground in the same time as it took to reach the maximumheight = 3.819 s
Total time taken willbe, T = 4.24 + 3.819 + 3.819 = 11.86
Velocity after thesecond collision
v3 =(9/10) (38.19)
v3 =34.37 m/s
Using the aboveinformation speed- time graph can be plotted
Question - 13 : - Provide clear explanations and examples to distinguish between:
( a ) The total length of a path covered by a particle and the magnitudeof displacement over the same interval of time.
( b ) The magnitude of average velocity over an interval of time, and theaverage speed over the same interval. [Average speed of a particle over aninterval of time is defined as the total path length divided by the timeinterval].
In ( a ) and ( b ) compare and find which among the two quantity isgreater.
When can the given quantities be equal? [For simplicity, considerone-dimensional motion only].
Answer - 13 : -
( a ) Let us consideran example of a football, it is passed to player B by player A and theninstantly kicked back to player A along the same path. Now, the magnitude ofdisplacement of the ball is 0 because it has returned to its initial position.However, the total length of the path covered by the ball = AB +BA = 2AB. Hence,it is clear that the first quantity is greater than the second.
( b ) Taking the aboveexample, let us assume that football takes t seconds to cover the totaldistance. Then,
The magnitude of theaverage velocity of the ball over time interval t = Magnitude ofdisplacement/time interval
= 0 / t = 0.
The average speed ofthe ball over the same interval = total length of the path/time interval
= 2AB/t
Thus, the secondquantity is greater than the first.
The above quantitiesare equal if the ball moves only in one direction from one player to another(considering one-dimensional motion).
Question - 14 : - A man walks on a straight road from his home to a market 2.5 km away witha
speed of 5 km/h. Finding the marketclosed, he instantly turns and walks back
home with a speed of 7.5 km h–1.What is the
(a) Magnitude of average velocity, and
(b) Average speed of the man over theinterval of time (i) 0 to 30 min, (ii) 0 to
50 min, (iii) 0 to 40 min? [Note: Youwill appreciate from this exercise why it
is better to define average speed astotal path length divided by time, and not
as the magnitude of average velocity.You would not like to tell the tired man on
his return home that his average speedwas zero !]
Answer - 14 : -
Distance to the market= 2.5 km = 2500 m
Speed of the man walking to the market= 5 km/h = 5 x (5/18) = 1.388 m/s
Speed of the manwalking when he returns back home = 7.5 km/h = 7.5 x (5/18) = 2.08m/s
(a) Magnitude of theaverage speed is zero since the displacement is zero
(b)
(i) Time taken toreach the market = Distance/Speed = 2500/1.388 = 1800 seconds = 30 minutes
So, the average speedover 0 to 30 minutes is 5 km/h or 1.388 m/s
(ii) Time taken toreach back home = Distance/Speed = 2500/2.08 = 1200 seconds = 20 minutes
So, the average speedis
Average Speed over ainterval of 50 minutes= distance covered/time taken = (2500 + 2500)/3000 =5000/3000 = 5/3 = 1.66 m/s
= 6 km/h
(ii) Average speedover an interval of 0 – 40 minutes = distance covered/ time taken = (2500+1250)/2400 = 1.5625 seconds = 5.6 km/h
Question - 15 : - In Exercises 3.13 and 3.14, we have carefully distinguished betweenaverage speed and magnitude of average velocity. No such distinction isnecessary when we
consider the instantaneous speed andthe magnitude of velocity. The instantaneous speed is always equal to themagnitude of instantaneous velocity. Why?
Answer - 15 : -
Instantaneous velocityand instantaneous speed are equal for a small interval of time because themagnitude of the displacement is effectively equal to the distance travelled bythe particle.
Question - 16 : - Look at the graphs (a) to (d) carefully and state, with reasons, which ofthese cannot possibly represent the one-dimensional motion of a particle.
Answer - 16 : -
None of the fourgraphs shows a one-dimensional motion.
(a) Shows twopositions at the same time, which is not possible.
(b) A particle cannothave velocity in two directions at the same time
(c) Graph showsnegative speed, which is impossible. Speed is always positive
(d) Path lengthdecreases in the graph, this is also not possible
Question - 17 : - The figure shows the x-t plot of the one-dimensional motion of aparticle. Is it correct to say from the graph shows that the particle moves ina straight line for t < 0 and on a parabolic path for t >0? If not,suggest a suitable physical context for this graph.
Answer - 17 : -
It is not correct tosay that the particle moves in a straight line for t < 0 (i.e., -ve) and ona parabolic path for t > 0 (i.e., + ve) because the x-t graph does notrepresent the path of the particle.
A suitable physicalcontext for the graph can be the particle is dropped from the top of a tower att =0.
Question - 18 : - A police van moving on a highway with a speed of 30 km h–1 firesa bullet at a thief’s car speeding away in the same direction with a speed of192 km h–1. If the muzzle speed of the bullet is 150 m s–1,with what speed does the bullet hit the thief’s car? (Note: Obtain that speedwhich is relevant for damaging the thief’s car).
Answer - 18 : -
Speed of the policevan = 30 km/h = 30 x (5/18) = 25/3 m/s
Speed of a thief’s car= 192 km/h = 192 x (5/18) = 160/3 m/s
Muzzle Speed of the bullet = 150 m/s
Speed of the bullet =speed of the police van + muzzle speed of the bullet
= (25/3)+ 150 = 475/3m/s
The relative velocityof the bullet w.r.t the thief’s car is
v = Speed of thebullet – Speed of a thief’s car
= (475/3) – (160/3) =105 m/s
The bullet hits thethief’s car at a speed of 105 m/s
Question - 19 : - Suggest a suitable physical situation for each of the following graphs
Answer - 19 : -
(a)The graph issimilar to kicking a ball and then it hits the wall and rebounds with a reducedspeed. The ball then moves in the opposite direction and hits the opposite wallthat stops the ball.
(b) The graph is showing a continuous change in the velocity of the object andat some instant it losses some velocity. Therefore, it may represent asituation where a ball falls on the ground from a certain height and reboundswith a reduced speed.
(c) A cricket ball moving with a uniform speed hit by a bat for a veryshort time interval.
Question - 20 : - The following figure gives the x-t plot of a particle executingone-dimensional simple harmonic motion. (You will learn about this motion inmore detail in Chapter 14). Give the signs of position, velocity andacceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Answer - 20 : -
In S.H.M.,acceleration, a = – ω2 x , ω is the angular frequency—-(1)
(i) At t = 0.3 s, x < 0 i.e., Position is negative. Moreover, as x isbecoming more negative with time, it shows that velocity is negative (i.e., v< 0). However, using equa (1), acceleration will be positive.
(ii) At t = 1.2 s, Positions and velocity will be positive. Accelerationwill be negative.
(iii) At t = -1.2 s, Position, x is negative. Velocity and accelerationwill be positive.