DYNAMICS

312 Engineering Mechanies -

Example 39: In an Asian games event of 100 mrun, an

start to his maximum velocity in a distance of 4 mand atrunshletetheaccelerates
determine re(a)mainhiisngunidlormb
velocity. If the athlete completes the race in 10.4 8.
(b) his maximum velocity.
Solution : Representing the given data, v(m/s)
we get v-f graph as shown in figure Ex.39.
Distance travelled = Area under v-t graph max

For 0 StSt,
A, = X Vmax X t, = 4
8

For t, S 1s 10.4
Vmax
0 ti
A = (10.4 - 10.4 -tË 104
Put value of t, from t)vmax =96 ....ii)
equation (i) in Fig. Ex.39
equation (ii) to get
104 8
Vmax XVmaX = 96

(10.4Vmax - 8) = 96
Vmax = 10 m/s.
Hence t, = =
= 0.8 s
Initial acceleration = Slope of v-t Vmax
diagram from 0 to 0.8 s
dv V.
max Vo
a =
di t - 0
10-0 = 12.5 m/s
0.8
.:. (i) Initial acceleration a = 12.5 mn/s
...Ans.
(ii) Maximum velocity vmax = 10 m/s Ans.
Example 40 :
100 km/hr from InAtotraveling a
car uniform B for t distance
sec.
of 3
If brakes are km between points A and D, car
D with deceleration from 100 km/hr applied for 4 seconds 1s driven
to
uniform speed of 60 km/hr. Find the km/hr and it takes t
60 between B and C to
value seconds to move fromgivea
Solution :
Representing
v-t diagram, weuse the given data by
of time t. C0
tv (m/s)
Distance travelled = Area under y-t 100 x5 A
[Refer figure Ex.40] diagram 18
60 x
18
3000 = 100
x+0o0x418

+ 60 x 5
18

t =65.5 s Ans,

Fig. Ex 40
 Rectilinear Mötion 313
af Particle
Chp 1 Kinematics

length of a railway line. The

over 2 km exceed
repairs are being carned out over the repair track should not
Example 41 : Track train is 90 kmph and the speed decelerates uniformly from 90 knph to
maximum speed of the repair track i600 m.
approaching the acoelerates to its full speed in
36 kmph. The trainAffer traveling on the repair track it
36kmph in 200 n. lostdue fo track repair.
Determine the tine v (m/s)t
the given data by
Solution : Representing figure Ex.4I. 25
shown in
v-tdiagram as 90 x 5 = 25 m/s
speed =
Given: Maximum 18
= 10m/s
Minimum speed = 36 x 18 10
2000 + 1600 A, Ag t (s)
in = 200 +
Total length of journey D
which
speed is affected 3800m A B t
C tg
t,
different distances are
going on,
When repair isfollowing Fig. Ex. 41
intervals.
covered in the
For portion AB Area under AB = Change in position
s
200 .. = 11.428

For portion BC in position

Area under BC = Change 200 s
10t, = 2000 .. h =
For portion CD position
under CD = Change in
Area
s
.. ta = 91.428
x - I600 91.428 = 302.856 s
11.428 + 200 +
t ta =
Total time t = t, + t, required to cover 3800 m is
of 25 m/s time
with uniform speed 3800 = 152 s
25
152 = 150.856 s Ans.
Time lost = 302.856 -
v (m/s)
at 42
: The race car starts from rest and
Example 42 40
along a straight road until it reaches a
travels seconds as shown by yt
speed of 42 m/s in 50 30
by the race
graph. Determine the distance travelled
and a-t graph.
car in 50 seconds. Draw x-t 20
18
t = 0, v =0,
Solution:Given initial condition, at 10

For x-t diagram 30 40 50

10 20
position (Ax)
Area under v-t diagram = Change in Fig. Ex. 42(a)
[Refer figure Ex.42(a)]
 314 Engineering Mechanios - DYNAMIcs

For 0sts 20s A, x 20 x 18 X0 -Xo 180 = X0 -0

2
For 20 stS30 s A, = 18 x 10 = Nå -X20 ’180 = X0 - 180

For 30 Sts 50 s A, 18 x20 +x 20 x 24 =Xs0 - X0 ’600 = Xsg -360

For a-t diagram
Slope of v-tdiagram = Acceleration [Refer figure Ex.42(a)]
dv V20- Vo 18 - 0 = 0.9 m/s
For 0s tS 20 s
dt 20-0 20 -0

For 20 StS 30 s a = dv V30- V20 18- 18

= 0
dt 30 - 20 30 - 20

For 30 stS 50 s a = dv Vs0 - V30 42 - 18

dt 50 - 30
= 1.2 m/s
20
Distance travelled by race car in 50 s is s = 960 m Ans.
x-t diagram and a-t diagrams are plotted as
shown in figure Ex.42(b) and figure Ex.42lci
x (m)
a(m/s)
1000
960 1.5
800
1.2
1.2
600
0.9 0.9
400
Straight line 0.6
200 360
180 0.3

10 20 30 40 50 t (s)
Fig. Ex. 42 (b) : -t 10 20
diagram 30 40 50 t(8
Fig. Ex.42(c) : a-t
JExample 43 : A car travels along a diagram
road with the speed shown by the straight v(m/s)
Determine the total v-t graph.
distance
until it stops when t= 48 8. the car travels 6

and a-t graphs. Also plot the x-!

4

Solution: Given initial condition, at t= 0, Vo v=--48

0, 2
For x-tdiagram
Area under v-t diagram =
[Refer figure Ex. 43(a)] Change in position (Ax) 10
20 30 40 48 tis
Fig. Ex. 43(a)
 Chp 1 Kinasies of Parttele Bactatneas Motion 315

For 0sts 30 s x 30 x 6 0- Xo Xa0 90 m

For 30 stS48 s A, x 18 x 6 Xs0 - X30 54 Xs - 90 Xn 144 n

Distance travelled by thecar in 48 sbefore it comes to rest 144 m Ans.

For a-t diagram
Slope of v-t diagram = Acceleration [Refer figure Ex. 43(a)]
For 0s tS30 s dv - 0 . 2 m/s
dt

For 30siS48 s a = dvdt = -0.333 m/s

I-t iagram and a-tdiagram are plotted as shown in figure Ex.43(b) and fc).
a(m/s

0.2 0.2
x (m)

144 0.1
150
48
t (s)
100 90 10 20 30 40 50

-0.1
50
t(s) -0.2
10 20 30 40 48
-0.3
-0.33
Fig. Ex. 43 (b) : x-t diagram
-0.4 Fig. Ex. 43(c) : a-t diagram

v (m/s)
Example 44 : Thev-t diagram for a particle
moving along straight line is shown in figure
Ex. 44 (a). Knowing thatx=-10 m at t =0 4
(a) Plot X-t and a-t diagram for 0<|< 50 s.
(b) Determine the maximum value of 2
position coordinate and the value of r for
which the particle is at a distance of 55 m
from the origin. 5 10 15 20 25 30 35 40 45 50 ts)

-2
Solution : Given initial condition, at t =0,
Xo =-10 m, vo = 2 m/s 4

(a) For x-t diagram -5

Area under v-t diagram = Change in position (Ar) Fig. Ex, 44(a) v-t diagram
[Refer figure Ex. 44 (a)]
316 Engineeriag Mechanicg DYNAMICS

For 0 stS I5 s A, = 15 x 2 =|s - o 30 = xs -(-10)

For 15 srs25 s A, -| 10 = X - X15 35 = x2s - 20

For 25 St s 30 s A;=x5x 5 = X30 -Xs 12.5 = X0 - 55

X30 62
For 30 sIS35s A, = -2 x 5x 5 = Xs -X0 ’ -12.5 = Xs - 67.5 S

For 35 stS 50s A =-15 x 5 = Ns0 - X35 -75 = Xs0 - 55 Xs0

. Maximum position coordinate is x= 67.5 m at t= 30 s Ans.
Att= 25 s and =35 s the particle is at a distance of 55 m.. Ans.
(b) For a-t diagram
Slope of v-t diagram = Acceleration (Refer figure Ex. 44(a)]
dv Vs - Vo
For 0S t< 15 s a =
dt 15 -0 2-2
15
= 0

For 15 stS 25 s a= dv V25 - V15 5-2 = 0.3 m/s?

dt 25- 15 10

For 25 stS35 s a = dv V35 - V2s -5- 5 = -l m/s

dt 35- 25 10
For 35 stS 50 s a = dv Vs0- V35 -5 -(-5) = 0
dt 50 - 35 15
x-t diagram and a-t diagram are plotted as shown in
figure Ex.44(b) and (c).
* (m)
a (m/s²)
70 67.5
60 0.4
55) 55 0.3 0.3
50
0.2
40
a =0 a=0
30
20 5 10 15 20 25 30 35 40 45 50 (8)
20
-0.2
10

0 -0.4
5 10 15 20 25 30 35 40 45\ 50 t (s)
-10
-20
-0.6.

-30
-20
-0.8
Fig. Ex. 44(b) : x-t diagram
-1

Fig. Ex. f4(c) : a-f diagram

 Chp 1 inonsatios of Partiole Roctiltnear Motion 317

ample 45: For a particle performing rectilineur

motion r-4 diagram is shown in flgure Ex45(a)
v(m/s)

Draw a-t diagram and -t diagram for motion if at

r2 5. A 20 m. What is the displacement during
b-10s2Also find the total distance travell ed

Solution : For a-t diagram

Slope of v-t diagram = Acceleration
[Refer figure Ex.45(a)] 8
dv V4 - Vo 8-0
For 0StS4s a= Fig. Ex.45(a) : V-t diagram
dt 4-0
a = 2 m/s v (m/s)
dv V6-V4 8-8
For 4StS6s a= =
= 0
dt 6-4 2

For 6 S tS 10 s a=
dv V10- V6-8-8-4 m/s
dt 10 -6 4
’t(s)
For x-t diagram 2 4

From the given condition at t = 2 s, x = x = 20 m, we Fig. Ex.46(b)

calculate xo at t = 0. We know that area under v-t
diagram gives change in position of the particle. By comparing two triangles shown in figure
Ex.53(b), we get
.. V = 4 m/s at t = 2 s

For 0<tS2s Area, A= x 2 x 4= X, - X0 4 = 20- X0 .Xo = l6 m

.. The particle starts from a position x = l6 m from the origin.
Nowcalculate the position of the particle at different intervals
16 = x4 - 16 X =32 m
For 0< tS4s A, =x4 x 8 = X4 - Xo
16 = Xs -32 ..X = 48 m
For 4st<6s A, = 2 x 8 = X6 - X4
8= X- 48 Xg= 56 m
For 6 <tS8s A, = x 2 x 8 = xy -Xo
o= 48 m
For 8<ts10s A= - x 2 x 8 = X10- Xg ’-8 = X0- 56
Now displacement during 6 to 10 s = A,+ A =8 - 8 = 0 Ans.

ial distance travelled = 4|+4 + 4s| + 4| = 16 - 16 - 8-8 = 48 m Ans.

 318 Engineering Mechanics DYNAMICS

a-t diagram and x-t diagram are drawn as x(m)

shown in figure Ex.45(c)and (d). 60 56
VË 80
50 48 30
a (m/s) -48
Straight line Thus
40
At s =
2 2 32,
30

0 t (s) At s
4 6 10 20
16
-2 10 At s

0
2 4 6 8 10 t(
am
Fig. Ex. 45 (c) : a-t diagram Fig. Ex. 45(d) : *-t diagram is th
partic
speed
AV(m/s)
Example 46 : The v-s graph of a rectilinear the p
moving particle is shown in figure Ex.46(a). 6
Find the acceleration of the particle at s = 20 m. Solut
80 m and 200 m.
4
Area
Solution: We have (Refe
dv 2
Acceleration, a = v
ds For 0

Hence calculate v and when s 20 m, 120 180210 240 s (m For 2

ds 60
80 m and 200 mn.
Fig. Ex. 46(a) For 4
V60 - Vo
For 0 S ss 60 m Slope dv
ds 60 -0 To fi
6-0
= 0.1
60
v (m/s) For (
dv V180 - V60
For 60 sss 180 m Slope ds 180- 60 6
6-6
= 0 For
120 4

dv V210 - V180
For 180 Sss 210 m Slope ds 210 - 180 2
0-6 V200 For
= -0.2 V20
30
180 210 sD
0
20 60 80 120 200
By comparison of triangles (Refer figure Ex. 46(b)] Fig. Ex.46(b) Maxi
V20 V60 V20 6
20 60
.. V20 = 2m/s when s = 20 m Maxi
20 60
Vgn = 6 m/s when s = 80 m
 Ch 1 Kinsmatics of Particie Rectillne at Motion 319

V|80 V200 V200 6

10 10 30 V'2002m/s when s 200 m

30
Thus we have

Ats = 20m. V=2 m/s and dv -0.1 . a= v 2 x0.1 - 0.2 m/s Ans.
ds ds
dv dv
At s = 80 m, V=6m/s and =0 Ans.
ds ds

Ats = 200 m, v=2m/s and dv =-0,2 .. a= v =2x (-0.2) = -0.4 m/s Ans.
ds ds

Vrample 47 : Figure Ex.47(a) shows a plot of a(m/s)

a v's rfor a particle moving along x-axis. What 12!.
is the speed and distance covered by the
particle after 50 sec ? Find also the maximum 8
speed and time at which the speed attained by
the particle. Draw v-tand x-t diagram.

Solution: Initial condition(assumed), at t = 0,

Xo =0, Vo =0 20 40 50 t (s)
Area under a-t diagram = Change in velocity (Av) Fig. Ex.47(a)
(Refer figure Ex.47(a)]
For 0<ts 20 s A= x 20 x 12 = 120 = V20 - Vo = V20 -0 .. V20 = 120ms
2

For 20 sts 40 s A, = 20 x 12 = 240 = V40 - V20 = V40 - 120 .V40 = 360 m/s
m)

x 10x 12 = 60 = V50 - 40 = Vs0 -360 .. Vso = 420 m/s

For 40 s ts 50 s A3 = 2

To find position of the particle, we use A A Ag

G G G
For 0 Sts 20 s, X20 = xo + Vo XL(20 - 0) t A1 ti 0
20 40 50 t (s)
=0+ 0+ 120 x x 20 t,
3
= 800 m
For 0 S ts 40 s, X40 = X0 t V20 Xl40- 20) t A, l
800 + 120 x 20+ 240 x x 20 ty = x20
2
= 5600 m

For 0 s ts 50s, Xs0 = X40 t V40 Xs0 - 40) Fig. Ex. 47(b)
= 5600+ 360 x 10 +60x 3
x10 = 9600m
Maximum speed of the particle occurs at t= 50 s and vmax = 420 m's Ans.

Maximum distance covered by a particle is x = 9600 m ... Ans.

 DYNAMICS
320 Engineering Mechanics -

and (d)
Plot v-t and x-f diagram as shown in figure Ex.47(c)
v (m/s) x (m)
Parabolic 10000
500
gnd degree 420
8000
9200
400
360

300 Straight 6000

line 5600,
200 4000
Parabolic

2nd 120,
100 2000
degree 800

10 20 30 40 50 t (s) 10 20 30 40
50
Fig. Ex.47(c) : v-t diagram Fig. Ex.47(d) : x-t diagram
Alternative solution for x-t diagram. (using v-tdiagram)
Area under v-tdiagram = Change in position (Ar)
[Refer figure Ex.47(c)]
a xb
For 0S ts20 s. A, = Here a = 20, b =120 and n =2 (Degree)
20 x 120
= 800 = X0-Xo = X0 -0 X20 = 800 m
2+1
(120 + 360) x20 = 4800 = X40 - X20= X40
For 20 <tS 40s, A, = 800 xA0 = 5600
2
nab
For 40<tS 50 s, A, = 10 x 360 + n+1 Here n = 2, a = 10, b= 60

+2 x 10x 60
A = 3600 2+ 1
= 4000= Xs0 -40 = Xs0 - 5600
.. Xs0 = 9600 m
Plot x-t diagram as shown in figure Ex.47(d)
a (m/s)

JExanple 48: For the a-t diagram of particle

shown in figure Ex.48(a) draw v- and x-l.
diagram. Also calculate the velocity at the end of
3 sec and distance travelled in 4 sec. Assume that
particle starts from rest from origin
1 2 3

Solution : Initialcondition(assumed), at / =0.

Xo = 0, 'o =0
Area under a-t diagram = Change in velocity (Av) -1
[Refer figure Ex. 48(a)] Fig Ex 48(a) * a-t diagra
 Chp 1 Kinematios of Partiate Rectilinar Motion 321
For 0 stSls, Area 4|=x I x I Av = V - Vo .. 0.5 = V- 0 V=0.5 m/s
For 1 sts2 s, Area A,= x lx I = Av = V,
-v 0.5 = V,- 0.5 V2 = l m/s
For 2S ts3 s, Area A, =-x1 x I =
Av = V3 - V; : -0.5 = V -1 Va 0.5 m/s
For 3 StS4s. Area A4 - x l x 1 = Av = VË -
V3. -0.5 = Va -0.5 Va 0 m/s
To find position of the particle from a-t
diagram, we use
A, Az Ag Ag
G
(s) G G,
1 2 4 t(s)

i,=x 1=t,
Fig. Ex. 48 (b)

For 0 StSls, X1 = o t Vo; tA, , = 0 + 0 + 0.5 xxl = 0. 166 m

For 0 s ts2s, X = X t vy, t A, t, = 0.166+ 0.5 x 1+ 0.5 x x | = 1m
For 0 <ts3s, x} = X, t vylg t A3 l; = 1+ 1x 1+ (-0.5) x =1.833 m
For 0 StS4 s, x4 = X; + V3 t4 t A4 t4 = 1.833 + 0.5 x 1+ (-0.5) x 3 = 2 m
Plot v-t and x-t diagram as shown in figure Ex.48(c) and (d).
v (m/s) x (m)
2.0+ 1.833 2.0

1.5

0.5 1.0+

0.5
0.166
A A4
1 2 3 4 t(s) 1 2 3 4 t (s)
Fig. Ex.48(c) : v-t diagram Fig. Ex. 48(d) : *-t diagram

Aiternative solution for x-tdiagram. (Using v-tdiagram)

Area under v-tdiagram = Change in position (Ar) [Refer figure Ex.48(c))
322 Engineering Mechnnice DYNAMICS

axh
For 0 stSls, Area 4, Here n2, u I , b=0.5

Ix0.5 -0.166 X- Xo -0
A
2+1
nab
For l stS 2s, Area A, Ix 0.5 + Here n= 2, a = 1, h= ).5
n+1
2 x Ix 0.5 =0.834 = X) -X = X, - 0.166
. A, = 0.5 + 2+1
nab
For 2sS3s. Area 4, = Ix 0.5 + n+1 Here n =2, a = 1, b= 0.5

.. A; =0.5 + 2x Ix 0.5 = 0.833= X} - X)

2+1 }-1X L331.
ab
For 3 <ts4s, Area A, = n +1
Here n= 2, a = 1, b=0.5

.: A4 = 1x 0.5 = Ax = X4 - X X-1.833
2+1 .X4 = 2m
Now, plot x-t diagram as shown in figure Ex. 48(d)
a(m/s)
2
Example 49 : The acceleration - time diagram
for the linear notion iS shown in figure
Ex.49(a). Construct velocity - time diagram
and displacement - time diagram for the motion
assuming that the motion starts with initial
velocity of 5 m's from the starting point.
Solution: Initial condition : At t= 0, Xo =0,
Vo = 5 m/s 12 t(s
Fig. Ex. 49(a) :a-t diagram
Area under a-t diagram = Change in velocity (Av)
[Refer figure Ex.49(a)]
For 0<tS6s, Area A, =x6x l =3 = V% - Vo = V6- 5 .. V = 8 ms
2

For 6 S t<12 s, Area A, = x6x 2 = 6= V12 - V6 = V2-8 . Vy2 = 14 m s

To find position of the particle for x-t diagram
A
(using a-t diagram), we use A, F
A
For 0< t<6 S, 0
12 t8 0
2 6 10
*6 = 0+5 x 6 + 3 x 4 5
= 42 m 10
For 0 <ts12 s. x2 = X6t V6 lz t Az ly i =x6 =4 20
.. X= 42 + 8 x6 + 6 x 2 t, =x 6 =2 Her
- 102 m Fig. Ex 49(b)
 Chp 1: Kinematics af Paticie Rectiline ar Motion 323
Plot v-fand x-tdiagram as shown in figure Ex.
v (m/s) 49(c) and (d).
x (m)
15
14
100 102
12

75
Degree 3
cubic
50
5 42

25

6 12 t (s) 6 12 t (s)
Fig. Ex.49(c) : v-t diagram Fig. Ex. 49 (d) : x-t diagram
Alternative solution for x-t diagram. (Using v-t diagram)
Area under v-t diagram = Change in position (Ar) (Refer figure Ex.49(c))
nab
For 0 <tS6s, Area A, = A + A = 5x6+ Here n = 2. a = 6, b = 3
n +1
.A, = 30 + O = 42 = X -Xo = X6 - 0 .. X = 42 m
2+1
ab
For 6 St s12 s, Area A, = A} + A = 6x 8 + n+1 Here n= 2, a= 6, b = 6
6x6
.. A, = 48 + = 102 n
2+ 1 = 60 = X2-X6 = X12 - 42
Now, plot x-tdiagram as shown in figure Ex. 49(d).
Example 50 : An airplane lands on the straight a (m/s)
runway, originally traveling at 20.5 m/s. If it is 5 10 15 20
subjected to the decelerations as shown in t (s)
figure Ex. 50(a), determine the time needed to
-0.5
stop the airplane. Construct v-t and s-t graph.
Solution :Given initial condition :
Att= 0, X =0., Vo =20.5 m/s, a=0
.4
For drawing v-tgraph from a-t diagram, we use Fig. Ex.50 (a) : a-t diagram
Area under a-t diagram = Change in velocity (Av)
0Sts s. Area A = 0= Vs - Vo = V; - 20.5 .. V = 20.5 m/s
Area 4, = 5x-0.5 = -2.5 = v0 -Vs = Vi0- 20.5 .. Vo = 18 m/s
5Sts10 s.
10sts 20 s, Area A = 10 x -1.4 = -14 = V20 - Vo = V20 - 18 . V20 4 m/s

Area AA = tX-(0.5) = -0.5t = V,-V0 = 0- 4 . t = 8s

2051St',
Hence plane will come to stop at time t' = (20 + ) = 28s Ans.
324 Bngineertng Mechanics DYNAMICS

Now construct v-t graph as shown infigure Ex. S0(b)

For drawing v-t graph from v graph, we use
Area under v-t diagram Change in position (A) |Refer figure Ex 50b1
(m/s) x (m)

25 350
20.5
20 300 308 75
250
15

10 200 198.75/
5 150

100 102.5)
5 10 15 20 25 2830 t (s) 50
Fig. Ex.50(b) : v-t diagram
5 10 15
20 25 2830
For 0 <tS5 s, Fig. Ex. 50(c) : x-t
Area A, = 5 x20.5 = x, - Xo = Xs diagram
-0
For 5sts 10 s, Area A, =20.5 +18)
2 x 5=X10 - Xs =X10 -
02.5
For 10 <IS 20 s, Area A; =
x 10 = X20 - X10 =
X0 -
198.75 X20 = 308. 7
For 20 <tS 28 s, Area A =
x 8x 4= Xg- X0 =
Now, construct x-t diagram as X2g - 308.75 Ig = 324. 7%
shown in figure Ex. 50(c)
Example 51: The acceleration for the
plane traveling along a straight small a (m/s)
in figure Ex.51 (a). runway is shown
50 m/s when Knowing that x=0 and y=
the plane at t=0, determine (a)
= 15 s, 25 s and 30 s the velocity of
(b) its
velocity during the interval 5 <t<25 s. average
Solution:Given initial
At t=0, X= 0, VÍ = condition : 10
A5 25 30
50 m/s 20
(a) Area under a-t
diagram = Change in velocity
(Refer figure Ex.51 (a)] (Av)
-1
0sts5 s, Area A=0 = Vs - Fig. Ex. 51 (a) : a-t diagran
Vo = Vs- 50
5StsI5 s, Area A, = x 10 x 1 =Vs
.. V7 50 ms
- Vs = Vs - 50
15StS 25 s, Area A, = x10 x 1 =
. Vs = 45 ms
2 Vs - Vis = Vs - 45 ..Vs= 50 ms
 Chp 1 Ginemacies of Patiele Rectslinear Moton 325

25 Sts 30s. Area 4, 0 V 0 -Vs V0-50 V10 50 m/s

. Velocity of plane at 15sis Vs45 m/s Ans.
I 2 5 sis Vs 50 m/s Ans.
75 30 s isV 5 0 m/s Ans.

(b) To find average velocity during the interval 5<< 25 s

Vo 50
Findx and xg from a-t diagram A A
G,
-0+ 50 x 5 +0 = 250m 30 t (s)
5 10 15 20 25
=NÍ t VÍ tAt
Here A
7 = A, i, t A_l;
= 0+ 50x 25 + (-5) x 15 + 5 x 5 Here t =25 s, t, = 15 s, ta= 5s
. X2s = 1200 m Fig. Ex. 51(b)
1200 - 250
Average velocity = Ax
At 25 - 5 25 - 5
= 47.5 m/s Ans.