| column A | Column B |
|---|---|
| The Slope of the speed-time graph is called | Average Velocity |
| The arithmetic mean of initial and final velocity | speed |
| What quantity is obtained by the area under speed time graph | acceleration |
| The slope of the distance time graph | distance |
Answer
column A
Column B
The Slope of the speed-time graph is called
acceleration
The arithmetic mean of initial and final velocity
Average Velocity
What quantity is obtained by the area under speed time graph
distance
The slope of the distance time graph
speed
|
Column A |
Column B |
|
Speed |
Scalar |
|
Distance |
Vector |
|
Displacement |
|
|
Acceleration |
|
|
Velocity |
|
Scalar : Speed ,distanceAnswer
Vector: Velocity,Acceleration ,displacement
|
Column A |
Column B |
|
22 km/h |
.266 m/s |
|
100 Km/h |
6.11 m/s |
|
10 m/min |
27.77 m/s |
|
2 km/s |
83.33 m/s |
|
5 km/min |
2000m/s |
22 km/h=6.11 m/sAnswer
100 Km/h=27.77 m/s
10 m/min= .266 m/s
2 km/s=2000 m/s
5 km/min=83.33 m/s
| Time | 2 sec | ? | ? | 10 min |
| Displacement | ? | ? | 25 m | ? |
| Velocity | ? | 10 m/s | ? | ? |
Answer
Time
2 sec
5 sec
5 sec
10 min
Displacement
4 m
25m
25 m
14.4 km
Velocity
4m/s
10 m/s
10 m/s
240m/s
Given, Answer
the initial velocity = 0 m /s,
acceleration = 3.20 m/s2,
t = 32.8 s
This is quite a basic question of 2nd equation of motion
$s=ut + \frac {1}{2}a t^2$
s = 1/2 at2
s=(.5)(3.20)(32.8)2= 1721.344m
Use 2nd equation of motionAnswer
$s=ut + \frac {1}{2}a t^2$
110 = [0.5a × (5.21)22]
a = 110 / (5.2122 × 0.5)
a = 8.10m/s2
Using 1st equationAnswer
v=u+at
v=gt where g =acceleration due to gravity=9.8 m/s2
v=9.8X 2.6=25.48 m/s
Using 2nd equation
$s=ut + \frac {1}{2}a t^2$
$s=\frac {1}{2}g t^2$
s=33.124m
$a =\frac {(v-u)}{t}$Answer
$a=\frac {(46.1-18.5)}{2.47}$
a=11.47m/s2
Here u=0, a=1.67m/s2 , s=1.40 , t=?Answer
Using 2nd equation
$s=ut + \frac {1}{2}a t^2$
t= 1.29 s
Here u=0, v=444m/s ,t=1.8Answer
$a= \frac {v-u}{t} = \frac {444}{1.8}=246.66 m/s^2$
Now
$s=ut + \frac {1}{2}a t^2$
s=399.6 m
Here v=7.10 m/s , s=35.4 m ,u=0 ,a=?Answer
Now,
$v^2 =u^2 + 2as$
a=.712 m/s2
Here v=65 m/s , s=? ,u=0 ,a=3m/s2Answer
Now,
$v^2 =u^2 + 2as$
distance= 704m
u = 22.4 m/s,v = 0, t = 2.55 sAnswer
using First equation of motion
v = u+at
0 = 22.4+aX2.55
a = -8.784 m/s2
Now using third equation of motion
$v^2=u^2 +2as$
0=(22.4)2 -2X8.784Xs
s=28.56m
Here v=0 , s=2.62m ,u=? ,a=-9.8 m/s2Answer
Now,
$v^2 =u^2 + 2as$
u= 7.17m/s
Here v=0 , s=1.29m ,u=? ,a=-9.8 m/s2Answer
Now,
$v^2 =u^2 + 2as$
u= 5.03m/s.
Now time going Up,
v=u+at
or
t=-u/a = 5.03/9.8 =.513 s
Total hang time = 2 * .513 =1.03s
Here v=521 m/s , s=.840m ,u=0 ,a=-?Answer
Now,
$v^2 =u^2 + 2as$
$a = 162 \times 10^3$ m/s2
Time to rise = 6.25/2= 3.125 s
s=? ,a =9.8 m/s2
Now,Answer
$s= \frac {1}{2} at^2$
s= 47.9m
s=370m ,a =9.8 m/s2
Now,Answer
$s= \frac {1}{2} at^2$
t= 8.69s
u=367 m/s ,a =? , s=.0621 m,v=0Answer
Now,
$v^2 =u^2 + 2as$
$a = -1.08 \times 10^6$ m/s2 (Negative sign shows deceleration)
depth of well= 57.0 mAnswer
speed= 47.6 m/sAnswer
acceleration = 2.86 m/s2Answer
time= 30.8s
acceleration = 15.8 m/s2Answer
v=0 (at maximum height),u=?,h=91.5m ,g=9.8m/s2Answer
From 3rd equation of motion
$v^2-u^2=2gh$
Substituting and solving, we get
$u=\sqrt (2gh)$
=42.34m/s
=94.43 miles/hr
a=0.2 m /s2, t=2 min=120 s, u=0Answer
$v=u+at$
v=24 m/s
$s=ut+ \frac {1}{2}at^2$
s=1440 m

a) AB and DEAnswer
b) C
c) OA and BC
d) CD
a. Odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.
b. Average speed is equal to the magnitude of average velocity When an object moves along a straight line in the same direction. In this case its total path length is the magnitude of displacement. Answer
a. DisplacementAnswer
b. v=72 km/hr =20 m/s , u=0 ,t=5 min=360 sec
$v=u+at$
a= .05 m/s2
$s=ut+\frac {1}{2}at^2$
s=3240 m
This Important Questions and answers on Motion for Class 9 physics with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.