The document analyzes the motion of a family car based on its velocity-time (v-t) graph to approximate the acceleration-time (a-t) and displacement-time (s-t) graphs. It shows that for the first 40 seconds, the car moves with constant velocity of 10 m/s and 0 acceleration, so displacement is simply velocity times time. From 40-80 seconds, the acceleration is calculated to be -0.25 m/s^2 based on the slope of the v-t graph, and formulas are used to determine the relationships between displacement, velocity, acceleration and time over this period. Tables of values for time, displacement, acceleration and velocity are also included.

1. ABSTRACT:
To analyze the motion of a family car as projected in the v-t graph and then approximate the a-t,
s-t graphs out of it. This can be the most common driving experience just plotted for a short time span.
ANALYSIS:
 Vdv=ads

푑푠
푑푡
=v

푑푣
푑푡
=a
 But here we are not given any relation between motion parameters so we will use the slope
determination technique to find the acceleration of the car, since slope of v-t graph is
acceleration.
1. Here the motion represents straight line for 0-40 sec of velocity i.e., 0 acceleration.
2. For 40-80 sec of velocity we have acceleration of:
푣2−푣1
푡2−푡1
0−10
80−40
=>
푚
푠2
=>-0.25=>
0-40 sec interval:
Velocity is constant at 10 m/s
푑푠
푑푡
푡
푡표 =∫ 푑푠
=v =>vdt=ds =>푣 ∗ ∫ 푑푡
푠
푠표 =>v*(t-t0) =s-s0 @s0=0 , t0=0
 The case of “0” acceleration, s=v*t.
 v*(t) =s , =>s=0.25*40 ,=>s=10m
 the car is just moving with constant velocity for 40 sec and equal distance of 10 meters is
covered every second.
0-80 sec interval:
Acceleration is constant = -0.25
풎
풔ퟐ
Vf=Vi+at, Vf=10-0.25*40, =>Vf=0

2. Vf=Vi-0.25*(t80-t40)
S=Vt*(t-t40)-0.125*(t-t40)2
S80=0*40-0.125*402=-200=200 m
 in this formula when we will be calculating the distance just for a first sight it will seems to give
first +ve increasing distance and then –ve distance , this is actually due to the made of this
formula which is based on the distance covered in the verge of velocity Vt at that time t (Vi*t) ,
and distance covered in the verge of acceleration or deceleration ,but if in case of deceleration
distance is –ve we take its mode (distance cannot be -ve),and adding them up in that case.
o So then the distance should come positive and increasing in concerned direction of
motion and for a variable motion scenario previous distance covered is to be added for
actual distance covered evaluation(in this final case distances direction can be taken
into account).
700
600
500
400
300
200
100
0
s-t
0 20 40 60 80 100
distance meters
time sec
Series1
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
a-t
0 20 40 60 80 100
accelerationm/s^2
time sec
Series1

3. ANNEXURE:
TABLE 1:
t s a v
0 0 0 10
1 10 0 10
2 20 0 10
3 30 0 10
4 40 0 10
5 50 0 10
6 60 0 10
7 70 0 10
8 80 0 10
9 90 0 10
10 100 0 10
11 110 0 10
12 120 0 10
13 130 0 10
14 140 0 10
15 150 0 10
16 160 0 10
17 170 0 10
18 180 0 10
19 190 0 10
20 200 0 10
21 210 0 10
22 220 0 10
23 230 0 10
24 240 0 10
25 250 0 10
26 260 0 10
27 270 0 10
28 280 0 10
29 290 0 10
30 300 0 10
31 310 0 10
32 320 0 10
33 330 0 10
34 340 0 10
35 350 0 10
36 360 0 10
37 370 0 10
38 380 0 10

4. 39 390 0 10
40 400 0 10
41 409.875 0.25 9.75
42 419.5 0.25 9.5
43 428.875 0.25 9.25
44 438 0.25 9
45 446.875 0.25 8.75
46 455.5 0.25 8.5
47 463.875 0.25 8.25
48 472 0.25 8
49 479.875 0.25 7.75
50 487.5 0.25 7.5
51 494.875 0.25 7.25
52 502 0.25 7
53 508.875 0.25 6.75
54 515.5 0.25 6.5
55 521.875 0.25 6.25
56 528 0.25 6
57 533.875 0.25 5.75
58 539.5 0.25 5.5
59 544.875 0.25 5.25
60 550 0.25 5
61 554.875 0.25 4.75
62 559.5 0.25 4.5
63 563.875 0.25 4.25
64 568 0.25 4
65 571.875 0.25 3.75
66 575.5 0.25 3.5
67 578.875 0.25 3.25
68 582 0.25 3
69 584.875 0.25 2.75
70 587.5 0.25 2.5
71 589.875 0.25 2.25
72 592 0.25 2
73 593.875 0.25 1.75
74 595.5 0.25 1.5
75 596.875 0.25 1.25
76 598 0.25 1
77 598.875 0.25 0.75
78 599.5 0.25 0.5
79 599.875 0.25 0.25
80 600 0.25 0