Motion Class 9 numericals

Some Numerical Of Motion

Hey there, here you will get numerical based on a motion for class 9. You will be known to solve the motion numerical problems here. You will get all science exercises here.

1. Graphical Representation of                                          

 Motion 

 Graphs provide a convenient method to present basic information about a variety of events

. For example, in the telecast of a 

 one-day cricket match. vertical bar graphs 

 show the run rate of a team in each over. As 

 you have studied in mathematics, a straight 

 line graph helps in solving a linear equation 

 having two variables. 

 To describe the motion of an object. we 

 can use line graphs. In this case, line graphs 

 show the dependence of one physical quantity. 

 such as distance or velocity, on another 

 quantity, such as time.

2. Describing Motion 

 We describe the location of an object by 

 specifying a reference point. Let us 

 understand this by an example. Let us 

 assume that a school in a village is 2 km north 

 of the railway station. We have specified the 

 position Of the school With respect to the 

 railway station. In this example, the railway 

 station is the reference point. We could have 

 also chosen other reference points according 

 to our convenience. therefore. to describe the 

 position of an object we need to specify a 

 reference point called the origin.

 

3, Rate of Change of Velocity 

 During uniform motion of an object along a 

 straight line, the velocity remains constant 

 With time. In this case, the change in velocity 

 Of the Object for any time interval is zero. 

 However, in non-uniform motion, velocity varies with time. It has different values at different instants

and at different points of 

 the path. Thus, the change in velocity of the object

 during any time interval is not zero. 

 Can we now express the change in velocity of an

 object? 

attain a velocity Of 6 m s-l in 30 s. Then he applies
brakes such that the velocity of

 the bicycle comes down to 4 m s in 

 the next 5 s. Calculate the acceleration 

 of the bicycle in both the cases.

Solution: 

 In the first case: 

 initial velocity, u = O ; 

 final velocity. v = 6 m s𑁦¹ ; 

 time, t 30 s . 

 From Eq. (8.3), we have 

a=(v-u) 
        t

Substituting the given values of u, v and in

the above equation. we get 

a= (6m s𑁦¹ 0m s𑁦¹) 
                30 s    

 = 0.2 m s𑁦² 

 In the second case: 

 initial velocity, u = 6 m s𑁦¹; 

 final velocity, v = 4 m s𑁦¹; 

 time, t = 5 s. 

 4m s=6m/s

 Then, a = (4 m s𑁦¹ - 6 m s𑁦¹)

                             5 s
                 = 0.4 m s²

 The acceleration of the bicycle in the 

 first case is 0.2 m/s² and in the second 

 the case, it is -0.4 m/s²

4.The odometer of a car reads 
 2000 km at the start of a trip and
 2400 km at the end of the trip. If the
 trip took 8 h. calculate the average
 the speed of the car in km h-l and m s-l.

Write CSS OR LESS and hit save. CTRL + SPACE for auto-complete.

Solution
Distance covered by the car.
s=2400 km  - 2000 km = 400 km

 

 =50 km h𑁦¹

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