Kinematics questions

Kinematics
For Competitive exams

Index
Rectilinear Motion
1) Introduction: rest and motion, Frame of reference , distance/displacement
2) Speed and velocity
3) Acceleration
4) Non uniform acceleration
5) Graphs: distance/displacement-time graph, speed/velocity-time graph,
acceleration-time graph
Projectile motion
1. Introduction to projectile
2. Oblique projectile
3. Horizontal projectile
4. Projectile on a moving platform
5. Projectile on an inclined plane
6. Elastic collision of projectile
Relative motion
1) Introduction
2) River boat problems, Wind aero plane problems , Rain umbrella problems
3) Velocity of approach/separation, condition for collision , minimum distance

Level 1 ncert
Ans. Displacement of magnitude 1 km and direction 60° with the initial direction;
total path length = 1.5 km (third turn); null displacement vector; path length = 3
km sixth turn); 866 m, 30°, 4 km (eight turn)

Level 1
A car covers the 1st half of the distance between two places at a
speed of 40 km h-1 and the 2nd half at 60 km h-1. What is the
average speed of the car ?
48

Level 1
A non-stop bus goes from one station to another station with a speed of 54
km/h, the same bus returns from the second station to the first station with a
speed of 36 km/h. Find the average speed of the bus for the entire journey.
43.2

Level 1 ncert
Ans. 182.2 m s–1

Level 1 ncert
4.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of
the park, then cycles along the circumference, and returns to the centre along OQ as shown in
figure 4.24. If the round trip takes 10 min, what is the (a) net displacement, (b) average
velocity, and (c) average speed of the cyclist ?
Ans. (a) O; (b) O; (c) 21.4 km h–1

Level 1 ncert
Ans. (a) 49.3 km h–1 ; (b) 21.4 km h–1. No, the
average speed equals average velocity
magnitude only for a straight path

Level 1 ncert
3.06 m s–2 ; 11.4 s

Level 1ncert
2s, 24 m, 21.26 m s–1

Level 2
A balloon is rising with constant acceleration 2 m/sec2. Two stones are released
from the balloon at the interval of 2 sec. Find out the distance between the two
stones 1 sec. after the release of second stone.
48 m
NOTE:- As the particle is detached from the balloon it is having the same
velocity as that of balloon, but its acceleration is only due to gravity and is
equal to g.

Level 3
A body moving with a constant retardation travels 5.7 m and 3.9 m in the 6th and
9th second respectively. When will the body stop moving?
t = s15s
6.0
9


Level 1 ncert
(a) v(t) = (3.0i – 4.0j); a (t) = – 4.0j
(b) 8.54 m s-1, 70° with x-axis.

Level 1(dt)
From the following displacement time graph find out the velocity of a moving body

Level 1 ncert
Ans. (a) A ......B, (b) A....... B, (c) B..... A , (d)
Same, (e) B.....A.... once.

Level 1 ncert
Ans. x < 0, v < 0, a > 0 ; x > 0 , v > 0 , a < 0 ; x <
0, v > 0 , a > 0

Level 1 ncert
3.19 Suggest a suitable physical situation for each of the following graphs (fig. 3.22)_

Level 1(vt)
The velocity time graph of a body moving in a straight line is shown in the
figure. The displacement and distance travelled by the body in 6 sec are
respectively
(a) 8m,16m
(b) 16m,8m
(c) 16m,16m
(d) 8m,8m
(a)

Level 1(dt)
The diagram shows the displacement-time graph for a particle moving in a
straight line. The average velocity for the interval t = 0, t = 5 is
c

Level 2
A car accelerates from rest at a constant rate ‘‘ for sometime. After which
it decelerates at a constant rate  to come to rest. If the total time elapsed
is t second, evaluate
(a) the maximum velocity reached, and
(b) the total distance travelled.



t
vmax
 
 



22
1 2
max
t
tvs

Level 1
A particle is projected from point O with velocity u in a direction making an
angle  with the horizontal. At any instant its position is at point P at right angles
to the initial direction of projection. Its velocity at point P is
(b)

Level 2
A particle is projected with velocity vo = 100 m/s at an angle q = 30o with the
horizontal. Find
(a) velocity of the particle after 2 s.
(b) angle between initial velocity and the velocity after 2s
(c) the maximum height reached by the projectile
(d) horizontal range and time of flight of the projectile.
Y
vo
X
q cos-1(0.98) = 10.8o

Last question
(a)      jˆviˆvv tytxt


Where iˆ and jˆ are the unit vectors along +ve x and +ve y axis respectively.
      jˆauiˆtauv tyyxxt 

Here, ux = vo cosq = 50 3 m/s
ax = 0
uy = vosinq = 50m/s
ay = -g ( g acts downward)
    jˆgtsinviˆcosvv oot qq

)t(v

=  jˆ21050iˆ350  = jˆ30iˆ350  m/s
 s/m65.91vvv 2
y
2
x2 

(b) jˆ50i350vo 

  jˆ30iˆ350v sec2 

   900015007500v.v 2o 

If  is the angle between ov

and  s2v

.
Then cos  =  
  65.911000
9000
vv
v.v
s2o
s2o





  = cos-1
(0.98) = 10.8o
.
(c) ya2uv y
2
y
2
y 
at y = ymax, vy = 0
 0 -   max
22
0 yg2sinv q  ymax = m125
g2
sinv 22
0

q
(d) R =
g
2sinu2
q
= 1732 m

Level 2
A projectile is thrown with a speed of 100 m/s making an angle of 600 with the
horizontal. Find the time after which its inclination with the horizontal is 450 ?

Level 2
1. (B) 2. (B) 3. (A) 4. (A) 5. (D)

Level 2
A ball 4 s after the instant it was thrown from the ground passes through a point P,
and strikes the ground after 5 s from the instant it passes through the point P.
Assuming acceleration due to gravity to be 9.8 m/s2 find height of the point P
above the ground.

Level 3
A football is kicked off with an initial speed of 20m/s at a projection angle of 450. A receiver
on the goal line at a distance of 60 m away in the direction of the kick starts running to meet
the ball at that instant. What must be his speed if he is to catch the ball before it hits the
ground? [Take g = 10m/s2]
45
P Q
R
RECEIVER
20 m/s
s/m25

Level 1
The barrel of a gun and the target are at the same height. As soon as the gun is fired,
the target is also released. In which of the following cases, the bullet will not strike
the target
(a) Range of projectile is less than the initial distance between the gun and the target
(b) Range of projectile is more than the initial distance between the gun and the
target
(c) Range of projectile is equal to the initial distance between the gun and target
(d) Bullet will always strike the target
(a) Condition for hitting of bullet with target
initial distance between the gun and target
Range of projectile.

Level 1
Two bullets are fired simultaneously, horizontally and with different speeds
from the same place. Which bullet will hit the ground first
(a) The faster one (b) Depends on their mass
(c) The slower one (d) Both will reach simultaneously
(d)

Level 2
A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h meter high
and b met wide, the ball will just hit the edge of nth step if n equals to

Level 1
An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation
of 6 km towards a point directly above the target on the earth’s surface. At an
appropriate time, the pilot releases a ball so that it strikes the target at the earth.
The ball will appear to be falling
(a) On a parabolic path as seen by pilot in the plane
(b) Vertically along a straight path as seen by an observer on the ground near the
target
(c) On a parabolic path as seen by an observer on the ground near the target
(d) On a zig-zag path as seen by pilot in the plane
(c)

Level 2
A stationary person A throws a ball as shown with speed u = 20 m/sec at an angle of 45° with horizontal.
At the same instant, another person B, at a distance of 10 m from A starts running with constant
velocity and catches the ball at point C. The velocity of ball and velocity of person B always lie in
same vertical plane; also the vertical level of point of projection and point of catching the ball from
ground is same. Then the speed v of person B will be : (g = 10 m/s
2
, Neglect air friction)
fp=kkuql kj ,d fLFkj O; fä A {kSfrt l s45° dsdks.ki j u = 20 m/sec dhpky l s,d xsan dksQsdrkgSA
bl h{k.k A l s10 m ehVj dhnwjhi j fLFkr nwl jkO; fä B fu; r osx l sxfr djuki zkjEHkdjrkgSvkSj fcUnq
C i j xsan dksi dM+ysrk gSA xsan rFkk O; fä B nksuksadsosx l eku Å/okZ/kj ry esagSrFkk i z{ksi .k fcUnq,oa
O; fä B }kjk xsan dksi dM+st kusokysfcUnqC dh/kjkry l sÅ¡pkbZl eku gSrksO; fä B dhpky gksxh&
(g = 10 m/s
2
, ok; qdk?k"kZ.kux.; ekusa)
V
BA
u
C
10m
(1) 15 2 m/sec (2) 7.5 2 m/sec (3) 6 2 m/sec (4) 2.5 2 m/sec
2

Last question
Range of projectile
i z{ksI; dhi j kl
2
u sin2
g
q
= 40 m
Initial distances are i zkj fEHkd nwj hAC = 40 cm, BC = 30 m
time of flight (mM~M; udky) 
 
2usin
g
q
=
2 20 sin 45
10
  
=
40 1
10 2
 =
4
2
= 2 2 sec.
 In this same time, the person should catch the ball.
vr %mM~M; u dky dscj kcj l e; esaO; fä B xsan dksi dM+
ysxkA
 V × 2 2 = 30
 V =
30 15
2 2 2
 m/sec.
=
15 2
2
= 7.5 2 m/sec

Level 2
A man is standing on a truck which is moving with a constant horizontal acceleration a = 10m/s
2
. When
speed of the truck is 10 m/s, the man throws a ball with speed 5 2 m/s with respect to truck in the
direction as shown in the diagram, then one second after the projection of the ball find the
displacement of the ball as observed by the man. (g = 10 m/s
2
)
,d O; fDr Vªd i j [ kM+kgSA; g Vªd fu; r {kSfrt Roj.k a = 10 m/s
2
l sxfr dj jgkgSt c Vªd dhpky
10 m/s gSrc O; fDr Vªd dsl ki s{kxsan dks5 2 dsosx l sQsadrkgS] i z{ksi .kfn'kkfp=kesaçnf'kZr gSAO; fDr
}kjkçsf{kr xsan dkO; fDr dsl ki s{ki z{ksi .kds,d l Sd.Mi 'pkr~foLFkki u Kkr djksA(g = 10 m/s
2
)
(1) 10 m (2) 20 m (3) 15 m (4) Zero

Last solution
w.r.t. man O; fDr dsl ki s{k
S = ut +
1
2
at2
= 5 2 × 1 +
21
( 10 2) 1
2
 = 0.

Level 2
See answers from reso outlet

Level 3
A batsman hits a ball at a height of 1.22 m above the ground so that ball leaves the bat
an angle 45o with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m
from the position of the batsman. Will the ball clear the wall if its range is 106.68m. If
not where it will hit the floor. Take g = 10 m/s2
106.68 m
BA
x
yvo
45o
1.22 m

Last
Solution :  R(range) =
g
2sinv2
0 q
 Rg
2sin
Rg
v2
0 
q
 as q = 45o
.
 vo = Rg . . . (1)
Equation of trajectory
Y = xtan45o
- o
cosv
gx
452 22
0
2
= x -
2
1
2
2
Rg
gx
using (1)
106.68 m
BA
x
yvo
45o
1.22 m
Putting x = 97.53, we get
y = 97.53 -
  35.8
1068.106
53.9710
2



Hence height of the ball from the ground level is h= 8.35 + 1.22 = 9.577m.
as height of the wall is 7.31m so the ball will clear the wall.

Level 1
iii) Find their velocity of approach
15

Level 1
Object A and B both have speed of 10 m/s. A is moving towards East while B is
moving towards North starting from the same point as shown. Find velocity of A
relative to B

Level 1 ncert
1250 m (Hint : view the motion of B relative to
A)

Level 1 ncert
1 ms–2 (Hint : view the motion of B and C
relative to A)

Level 2
A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s
respectively (g = 10 m/s2. Find separation between them after half second,
one seconds,1.5 seconds, two seconds, three seconds. Assume ball doesn’t
rebound
2.5,5,2.5,0,0

Level 2
A ball is thrown downwards with a speed of 20 m/s from the top of a building 150
m high and simultaneously another ball is thrown vertically upwards with a speed
of 30 m/s from the foot of the building. Find the time after which both the balls will
meet. (g = 10 m/s2)

Level 1ncert
Ans. 15 min, 750 m

Level 1 ncert
Ans. About 18° with the vertical, towards the
south.

Level 1 ncert
Ans. East (approximately)

Level 3
A swimmer crosses a river with minimum possible time 10 second. And when he
reaches the other end starts swimming in the direction towards the point from
where he started swimming. Keeping the direction fixed the swimmer crosses the
river in 15 sec. The ratio of speed of swimmer with respect to water and the
speed of river flow is (Assume constant speed of river & swimmer)

Level 2
A swimmer wants to cross a river to reach point B directly from A. The speed of the swimmer in still river
and that of river flow are same. The dotted line AN is normal to river flow direction. For swimmer to
reach point B, the angle his velocity relative to river will make with line AN is given by q. Then the
value of q is :
fp=kkuql kj , d rSjkd unhdksi kj djrsgq, A fcUnql sl h/ksB i j i gq¡prkgSArSjkd dhpky ¼i kuhdsl ki s{k½
rFkkunhi zokg dhpky , d l eku gSAfcUnqor js[ kkAN unhi zokg dsyEcor~gSAfcUnqB i j i gq¡pusdsfy,
rSjkd dkunhdsl ki s{kosx] js[ kkAN dsl kFkq dks.kcukrkgSrksq dkeku gksxk&
(1) 60° (2) 30° (3) 45° (4) 0°
u sin (q + 30°) = v sin 60°
q = 30°
u
30º
º

Level 2
A car is travelling north along a straight road at 50 km hr
–1
. An instrument in the car indicates that the
wind is directed towards east. If car’s speed is 80 km hr
–1
, then instrument indicates that the wind
is directed towards south – east. Then angle made by wind’s direction is given by.
(1) 1 3
tan NofE
5
  
q   
 
(2) 1 5
tan NofE
3
  
q   
 
(3) 1 1
tan Nof W
2
  
q   
 
(4) q = tan
–1
(5) N of E
b

A car is travelling north along................
, d d kj l h/khl M+d i j mÙkj fn' kk................
Sol. W GV
r
= x y
ˆ ˆV i V j
r
1W CV
r
is towards east
1W CV
r
i woZd hr j Q
Vx
Vy
50
 Vy = 50 m/s ; Vx > 0
2W CV
r
is towards south-east
2W CV
r
nf{k. k i woZd hr j Q
Vx
Vy
80
45°
 Vx = 30 m/s
50
30
1 5
tan Nof E
3
  
  
 
q

Level 2
A projectile is fired with velocity v0 from horizontal surface making an angle q with vertical. If the wind
imparts constant horizontal acceleration ‘a’ to the left then at what angle ‘q’ projectile must be fired
so that it returns to the point of projection. (Assume ‘g’ to be uniform)
{kSfr t l rg l s, d i z{ksI; dksv0 osx l sÅ/okZ/kj l sq dks.ki j i z{ksfi r fd; kt kr kgSA; fn ok; qckW; sr jQ , d
fu; r {kSfr t Roj.k ‘a’ i znku djr hgksr ksi z{ksi .k dks.k ‘q’ dk eku D; k gksuk pkfg, fd i z{ksI; oki l i z{ksi .k
fcUnqi j Vdjk; s(‘g’ dksl e: i ekfu; sA)
v0
a
(1) 1 a
tan
2g
  
  
 
q (2)
2
1
2
2a
tan
g
  
  
 
q (3) 1 a
tan
g
  
  
 
q (4) 1 g
tan
a
  
  
 
q
3

Last question
33. A projectile is fired with velocity v0................
{kSfr t l r g l s, d i z{ksI; dksv0 osx................
Sol.
2
y 0
1
S v cos t gt 0
2
 q   
02v cos
t
g

q
2
x 0
1
S v sin t at
2
 q  
02v sin
t
a

q
0 02v cos 2v sin
g a

q q
a
tan
g
q

Level 2
A man starts running along a straight road with uniform velocity ûi observes that the rain is falling vertically
downward. If he doubles his speed, he finds that the rain is coming at an angle q to the vertical.
The velocity of rain with respect to the ground is : ( î is unit vector along horizontal direction and ˆj
is unit vector vertically downward)
,d O; fDr fu; r osx ûi l sl h/khl M+d dsvuqfn'knkSM+uki zkjEHkdjrkgS] rksog ns[ krkgSfd cjl kr dhcawns
m/okZ/kj uhpsdhvksj fxj jghgSSA vxj og vi uhpky nqxuhdj ysrk gSrksml scjl kr dhcwansm/oZl s
qdks.ki j vkrhgqbZi zrhr gksrhgSAcjl kr dhcw¡nksadkt ehu dsl ki s{kosx gSA({kSfrt fn'kkdsvuqfn'kbdkbZ
l fn'kî rFkkÅ/okZ/kj uhpsdhfn'kkdsvuqfn'kbdkbZl fn'k ˆj gS)
(1) ˆ ûi utan j q (2) ˆ ûi ucot j q (3) ˆ ûi ucot j q (4)
u ˆ î uj
tan

q
2

Last solution
RMv
r
= R mv v
r r
RMv
r
= R
ˆv ui
r
= (VRX
î + VRY
ˆj ) – u î
Since RMv
r
, has only y component with respect to the man.
pwafd RMv
r
, dsfy , vknehd sl ki s{k d soy y ?kVd gSA
so, vr % VRX – u = 0
 VRX = u
After doubling the speed - pky nqxquhd j usd sckn
Rm´v
r
= (u î + vRy
ˆj ) – 2u î
= – u î +vRy
ˆj
given fn; k gStan q =
Ry
u
v
 vRy = u cot q
so bl fy , Rv
r
= u î + u cot q( ˆj )

Level 2
A ship is sailing due north with speed 9 m/s in sea. Wind is blowing with speed 2 m/s from S–W
direction. A person holding a flag in his hand running on the deck of ship with speed 3 m/s due
east relative to ship. Direction in which flag will flutter is :
(1) tan
–1
(2) W of N (2) tan
–1
(2) S of E
(3) tan
–1
(2) N of W (4) none of these
Vs = ˆ9 j
Vw =
ˆ ˆI j
2
2
 
 
  
= ˆ ˆI j
Vms = ˆ3I = Vm –Vs
Vm = ˆ3I +Vs = – ˆ ˆ3I 9 j
Vwm = Vw– Vm
= ˆ ˆ ˆ ˆ(i j) – [3i 9 j] 
= ˆ ˆ2i – 8 j 

Level 2
It is raining with a velocity of 5 2 m/s at an angle of 45° with vertical. To a man driving with constant
velocity ‘v’ down the inclined plane the rain appears to move vertically upwards with a velocity of 5
m/s. If q is the angle of the incline plane with the horizontal, then v and q are :
m/okZ/kj dsl kFk45° dsdks.ki j 5 2 m/s dsosx l sckjh'kgksjghgSA{kSfrt l s>qdko dks.kq okysur ry
i j uhpsdhvksj V fu; r osx l sxfr djrsO; fDr dksckjh'k5m/s dsosx l sm/okZ/kj Åi j dhvksj fxjrhgqbZ
i zrhr gksrhgSrksv rFkkq gS:
(1) 5 m/s , tan
–1
5 (2) 5 5 m/s, tan
–1
5 (3) 5 5 m/s, tan
–1 1
2
(4) 5 5 m/s, tan
–1
2
q
v
5 m/s
5 m/s
v cos q = 5
v sin q – 5 = 5
v sin q = 10

Level 2
A person standing on the bank of a river of width 1 km wants to cross the river in minimum possible time. If
the speed of the man in still water is
10
3
m/sec and speed of river flow is
10 m/sec, then find the distance travelled by the person with respect to ground when he reaches
the opposite bank of the river.
1 km pkSM+hunhds,d fdukjsi j [ kM+kgqvk,d O; fDr U; qure l aHko l e; esaunhdksi kj djukpkgrkgSA
; fn O; fDr dh' kkar i kuhesapky
10
3
m/sec rFkk unhdhpky 10 m/sec gksrks/kjkry dsl ki s{k O; fDr
}kjkr; dhxbZnwjhKkr djkst c og unhdsfoi fjr fdukjsi j i gqprkgS
(1) 4 km (2) 6 km (3) 3 km (4) 2 km
4) 2km

Last solution
A person standing on the bank of a................
1 km pkSM+hunhds, d fdukj si j ................
Sol. tan q =
u
v
=
10
10/ 3
= 3
u=10m/sSv
D=1km u
S =
D
cosq
=
1km
cos60
= 2km
Ans. 2

Level 3
Rain is falling with velocity ˆ ˆ4i 3j m/sec. A person starts moving with acceleration 2ˆ1im/sec on a
horizontal road. The difference between two time instants in seconds at which the rain appears to
fall at 45º with vertical to the person is : (x is horizontal and y is vertically upward direction)
ckfj'kosx ˆ ˆ4i 3j m/sec. l sfxj jghgSA,d O; fDr {kSfrt l M+d i j 2ˆ1im/sec Roj.kl sxfr i zkjEHkdjrk
gS] mu nks{k.kksadse/; l e; vUrjky D; kgksxkAft l l e; ckfj'kO; fDr dksm/okZ/kj l s45º dks.ki j fxjrh
gqbZi zrhr gksrhgS%(x {kSfrt rFkky m/okZ/kj Åi j dhrjQ gSA)
(1) 6 sec (2) 12 sec (3) 8 sec (4) 4 sec

Last solution
w.r.t. person O; fDr dsl ki s{k
± tan45º =
4 1t
3

4 1t
3

= ±1
t = 1,7
t = 7 – 1 = 6

Questions on Velocity of Approach /
Separation

Level 1
22,-22,22
iii) Find their velocity of approach.

Level 2
Particles A and B are moving with constant velocity in two perpendicular directions as shown in diagram at
t = 0, then the velocity of separation between A and B at t = 1 second is :
(1) 5 2 m/s (2)
7
2
m/s (3) 7 2 m/s (4)
5
2
m/s
At t = 1s i j , q = 45º,
Vsep = 3 cosq + 4sinq
=
1 1
3 4
2 2
 =
7
2

Level 2
A stone is projected with speed u aiming the mango at some height from ground as shown in figure. At the
instant of projection of stone the mango gets released (drops) from the tree. Then which of the
following is correct. (Distance between foot of tree and point of projection is less then the range of
projectile)
(1) The stone will hit the mango for all values of u.
(2) The stone will hit the mango only for a special value of u.
(3) The stone will never hit the mango
(4) None
Motion of stone with respect to mango is directly towards mango
So, they will collide.
vke dsl ki s{ki RFkj dhxfr vke dhvkSj gSA
vr %] ; g vke l sVdj k; sxkA

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