Chapter 3 Motion in a Straight Line Solutions
Question - 21 : - The figure gives the x-t plot of a particle in one-dimensional motion.Three different equal intervals of time are shown. In which interval is theaverage speed greatest, and in which is it the least? Give the sign of averagevelocity for each interval.
Answer - 21 : -
Interval 3 is thegreatest and 2 is the least. The average velocity is positive for interval 1and 2 and it is negative for interval 3.
Question - 22 : - The following figure gives a speed-time graph of a particle in motionalong a constant direction. Three equal intervals of time are shown. Inwhich interval is the average acceleration greatest in magnitude? In whichinterval is the average speed greatest? Choosing the positive direction as theconstant direction of motion, give the signs of v and a in the three intervals.What are the accelerations at points A, B, C and D?
Answer - 22 : -
The change in thespeed with time is maximum in interval 2. Therefore, the average accelerationis greatest in magnitude in interval 2.
The average speed ismaximum in interval 3
The sign of velocityis positive in interval 1, 2 and 3. The acceleration depends on the slope. Theacceleration is positive in interval 1 and interval 3 as the slope is positive.The acceleration is negative in interval 2 as the slope is negative.
Acceleration at A, B,C and D is zero since the slope is parallel to the time axis at these instants.
Question - 23 : - A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on astraight road for 10 s, and then moves with uniform velocity. Plot the distancecovered by the vehicle during the nth second (n = 1,2,3….) versus n. What doyou expect this plot to be during accelerated motion: a straight line or aparabola?
Answer - 23 : -
For a straight line,the distance covered by a body in nth second is :
SN = u + a (2n – 1)/2 . . . . . . . . ( 1 )
Where,
a = Acceleration
u = Initialvelocity
n = Time = 1, 2, 3, . . . . . , n
In the above case,
a = 1 m/s2 and u = 0.
∴ SN =(2n – 1) / 2 . . . . . . . . . . .( 2 )
This relation shows that:
SN ∝ n .. . . .. . . . . ( 3 )
Now substitutingdifferent values of n in equation ( 2 ) we get:
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| SN | 0.5 | 1.5 | 2.5 | 3.5 | 4.5 | 5.5 | 6.5 | 7.5 | 8.5 |
This plot is expectedto be a straight line.
A boy standing on a stationary lift (open from above) throws a ballupwards with the maximum initial speed he can, equal to 49 m s-1.How much time does the ball take to return to his hands? If the lift startsmoving up with a uniform speed of 5 m s-1 and the boy againthrows the ball up with the maximum speed he can, how long does the ball taketo return to his hands?
Answer - 24 : -
Initial velocity ofthe ball, u = 49 m/s
Case : I
The boy throws theball upwards when the lift is stationary. The vertically upward direction istaken as the positive direction. The displacement of the ball is zero.
Considering theequation of motion
s=ut + (1/2)at2
0 = (49)t + (1/2)(-9.8)t2
t = (49 x 2)/9.8 =98/9.8 = 10 sec
Case : II
As the lift startsmoving with a speed of 5 m/s, the initial speed of the ball will be 49 m/s + 5m/s = 54 m/s
The displacement ofthe ball will be s = 5t’
Therefore, the timetaken can be calculated using the formula
s = ut + (1/2) at2
5t’ = (54) t’ +(1/2)(-9.8) t’2
t’ = 2(54 – 5)/9.8 =10 sec
The time taken willremain the same in both the cases.
Question - 25 : - On a long horizontally moving belt figure, a child runs to and fro with aspeed 9 km h–1 (with respect to the belt) between hisfather and mother located 50 m apart
on the moving belt. The belt moves witha speed of 4 km h–1. For an observer on a
stationary platform outside, what isthe
(a) speed of the child running in thedirection of motion of the belt ?.
(b) speed of the child running oppositeto the direction of motion of the belt?
(c) time taken by the child in (a) and(b)?
Which of the answers alter if motion isviewed by one of the parents?
Answer - 25 : -
Speed of child = 9 kmh-1
Speed of belt = 4 km h-1
(a) When the boy runs in the direction of motion of the belt, then his speed asobserved by the stationary observer = (9 + 4) km h-1 = 13 km h-1.
(b) When the boy runsopposite to the direction of motion of the belt, then speed of child asobserved by the stationary observer = (9 – 4) km h-1 = 5 km h-1
(c) Distancebetween the two parents = 50 m = 0.05 km
Speed of the boy asobserved by both the parents is 9 km h-1
Time taken by the boyto move towards one of the parents =0.05 km/9k h-1=0.0056 h =20 S
Question - 26 : - Two stones are thrown up simultaneously from the edge of a cliff 200 mhigh with initial speeds of 15 m s–1 and 30 m s–1. Verify that the graph shownin Fig. 3.27 correctly represents the time variation of the relative position ofthe second stone with respect to the first. Neglect air resistance and assumethat the stones do not rebound after hitting the ground. Take g = 10 m s–2.Give the equations for the linear and curved parts of the plot.
Answer - 26 : -
For the first stone:
Given,
Acceleration, a =–g = – 10 m/s2
Initial velocity, uI = 15 m/s
Now, we know
s1 = s0 + u1t +(1/2)at2
Given, height of the tree, s0 = 200 m
s1 = 200 + 15t – 5t2 . . .. . . . . . . ( 1 )
When this stone hits the jungle floor, s1 = 0
∴– 5t2 +15t + 200 = 0
t2 – 3t – 40 = 0
t2 – 8t + 5t – 40 = 0
t (t – 8) + 5 (t – 8) = 0
t = 8 s or t = – 5 s
Since, the stone was thown at time t = 0, the negative sign is notpossible
∴t = 8 s
For second stone:
Given,
Acceleration, a =– g = – 10 m/s2
Initial velocity, uII = 30 m/s
We know,
s2 = s0 + uIIt + (1/2)at2
= 200 + 30t – 5t2 . . . . . . . . . . . . . ( 2 )
when this stone hits the jungle floor; s2 = 0
– 5t2 + 30 t + 200 = 0
t2 – 6t – 40 = 0
t2 – 10t + 4t + 40 = 0
t (t – 10) + 4 (t – 10) = 0
t (t – 10) (t + 4) = 0
t = 10 s or t = – 4 s
Here again, the negative sign is not possible
∴ t = 10 s
Subtracting equations ( 1 ) from equation ( 2 ), we get
s2 – s1 = (200 + 30t -5t2) –(200 + 15t -5t2)
s2 – s1 =15t . . . . . . . . . . . . .. . . . . . ( 3 )
Equation ( 3 ) represents the linear trajectory of the two stone, because tothis linear relation between (s2 – s1)and t,, the projection is a straight line till 8 s.
The maximum distance between the two stones is at t = 8 s.
(s2 – s1)max = 15× 8 = 120 m
This value has been depicted correctly in the above graph.
After 8 s, only the second stone is in motion whose variation with time isgiven by the quadratic equation:
s2 – s1 = 200 + 30t – 5t2
Therefore, the equation of linear and curved path is given by :
s2 – s1 = 15t (Linear path)
s2 – s1 =200 + 30t – 5t2 (Curved path)
Question - 27 : - The speed-time graph of a particle moving along a fixed direction isshown in the figure. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s,
(b) t = 2 s to 6 s.
Answer - 27 : -
(a) Distance traversedby the particle between t = 0 s and t = 10 s
= area of the triangle= (1/2) x base x height
= (1/2) x 10 x x12 =60 m
Average speed of theparticle is 60 m/ 10 s = 6 m/s
(b) The distancetravelled by the particle between t = 2 s and t = 6 s
= Let S1 bethe distance travelled by the particle in time 2 to 5 s and S2 bethe distance travelled by the particle in time 5 to 6 s.
For the motion from 0sec to 5 sec
Now, u = 0 , t = 5 , v= 12 m/s
From the equation v =u + at we get
a = (v – u)/t = 12/ 5= 2.4 m/s2
Distance covered from2 to 5 s, S1 = distance covered in 5 sec – distance covered in 2 sec
= (1/2) a (5)2 –(1/2) a (2)2 = (1/2) x 2.4 x (25 – 4) = 1.2 x 21 = 25.2 m
For the motion from 5sec to 10 sec , u = 12 m/s and a = -2.4 m/s2
and t = 5 sec to t = 6sec means n = 1 for this motion
Distance covered inthe 6 the sec is S2 = u + (1/2) a (2n – 1)
= 12 – (2.4/2) (2 x 1– 1) = 10.8 m
Therefore, the totaldistance covered from t = 2 s to 6 s = S1 + S2
= 25.2 + 10.8 = 36 m
Question - 28 : - The velocity-time graph of a particle in one-dimensional motion is shownin the figure.
(a) x (t2) = x (t1) + v (t1) (t2 –t1) + (1/2) a(t2 – t1)2
(b) v(t2) = v(t1) + a(t2 – t1)
(c) aaverage = [ x(t2) – x (t1)]/(t2 – t1)
(d) aaverage = [ v(t2) – v (t1)]/(t2 – t1)
(e) x (t2) = x (t1) + v (t2 – t1)+ (1/2) aav (t2 – t1)2
(d) x(t2) – x (t1) = Area under the v-t curve bounded by t- axis and thedotted lines.
Answer - 28 : -
The graph has anon-uniform slope between the intervals t1 and t2 (sincethe graph is not a straight line). The equations (a), (b) and (e) does notdescribe the motion of the particle. Only the relations (c), (d) and (f) arecorrect.