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Understanding Hooke's Law: The Foundation of Elasticity| Chapter 5 Physics 9th

HOOKE'S LAW   Introduction In physics, Hooke's Law is one of the fundamental principles governing how objects deform under external forces . Named after the 17th-century British physicist Robert Hooke, this law provides a crucial understanding of the behavior of elastic materials, such as springs and rubber bands. Whether stretching a rubber band or compressing a spring, Hooke's Law helps explain what happens when forces act on these materials. What is Hooke's Law: Hooke's Law states that the force F needed to extend or compress a spring by some distance x is proportional to that distance. Mathematically, it is expressed as: F= -kx Here k represents the spring constant, which is the measure of the stiffness of the spring, and x is the displacement from the displacement position.  The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.  Understanding the Spring Constant: The spring constant k is a critical co

Some Solved Problems on Speed, Velocity and Acceleration Physics 9th

 SOME SOLVED PROBLEMS ON SPEED VELOCITY AND ACCELERATION





Hello students in this article I have posted few solved problems on Speed, Velocity and Acceleration. 

Example#01: A sprinter completes its 100 meters race in 12 sec. Find its average speed.
Solution:
            From the given data we have;
                        distance = 100 meters
                        Time taken = 12 seconds
                        Average speed=?
            As we know that 
                    Avg. speed= distance travelled/time taken
                    v= 100/12
                    v= 8.33 meters/sec
Example#02: A car travels 700 meters in 35 seconds. What is the speed of the car?
Solution:
            From the given data we have;
                Distance travelled by the car = 700m.
                   Time taken = 35 sec.
                    Speed = v= ?
            Speed= distance travelled/ time
            v= 700/35 = 20m/s.
The car travels at speed of 20 m/s.

 Example#03:The speed of a train is 108km/h. How much distance will be covered in 2 hours?
Solution:
                From the given data we have
                Speed of train = 108km/h = (108*1000)/3600 = 30m/s
                Time taken = 2 hours = 2*3600 = 7200 sec.
                distance =?
                Since,
                            Speed = distance/ time
                            Distance = speed*time
                             s= 30*7200 =2,16000 meters.

 Example#04:A cyclist completes  half round of a circular track of radius 318 m in 1.5 minutes. Find its speed and velocity. 
Solution:
                From given data
                    
Speed and velocity numericals
        Radius of the track=318 m
         Time taken = 1.5 minutes
                            = 90 seconds.
        speed and velocity=?
         Distance covered by the cyclist for a circular path can be given as;
                        s=pi*radius
                    Substituting the values we can get
                        s= 3.142*318 = 999m.
As we are asked to calculate both speed and velocity so we need to calculate both distance and displacement to calculate speed and velocity. As we have already calculated the distance now is displacement can also be calculated as;
                    displacement =2*radius
                    displacement= 2*318 =636m.
Now we have both values i.e. distance and displacement so now finally we can calculate speed and velocity as;
                    speed = distance/time
                    speed= 999/90 =11.1 m/s.
Velocity can be calculated as;
                    Velocity= displacement/time
                    Velocity = 636/90 =7.07m.

Example#05:A bus starts from rest and travels along the straight path and its velocity becomes 15 m/s in 5 seconds. Calculate the acceleration of the bus? 
Solution:
            From the given data we have;
                Vi= 0 m/s [As bus starts from rest]
                Vf= 15m/s
                t=  5 seconds.
                a=?
        since the formula to calculate acceleration is;
                a= (Vf-Vi)/ t
                a= (15 - 0)/5
                a= 3m/s^2

Example#06:A car starts from rest. Its velocity becomes 20 m/s in 8 s. Find its acceleration.
Solution:
                From the given data we have;
                        Since car is starting from rest so its initial velocity will be = Vi= 0m/s
                        And after 8 seconds which is time given it attains a velocity of 20 m/s which is final velocity of the car and we are asked to calculate the acceleration.
            a= (Vf-Vi)/t
            a= (0-20)/8 = 20/8
            a= 2.5m/s^2

Example#07:Find the retardation produced when a car is moving at a velocity of 30 m/s slows down uniformly  to 15m/s in 5 seconds.
Solution:
                From the given data;
                a= ? (retardation / negative acceleration)
                Vi = 30m/s
                Vf= 15m/s
                 t= 5 sec
                 a= (Vf-Vi)/t
                   a=(15-30)/5= -15/5
                    a= -3m/s
Since the value of acceleration is negative therefore it is called as Retardation/ Negative acceleration.






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