Page 95 - Understanding NCERT Science 09
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Activity _____________7.11                        If you carefully note, on being released
                                                               the stone moves along a straight line
                •    Take a piece of thread and tie a small    tangential to the circular path. This is
                     piece of stone at one of its ends. Move   because once the stone is released, it
                     the stone to describe a circular path     continues to move along the direction it has
                     with constant speed by holding the
                     thread at the other end, as shown in      been moving at that instant. This shows that
                     Fig. 7.9.                                 the direction of motion changed at every point
                                                               when the stone was moving along the circular
                                                               path.
                                                                  When an athlete throws a hammer or a
                                                               discus in a sports meet, he/she holds the
                                                               hammer or the discus in his/her hand and
                                                               gives it a circular motion by rotating his/
                                                               her own body. Once released in the desired
                Fig. 7.9:  A stone describing a circular path with
                                                               direction, the hammer or discus moves in
                          a velocity of constant magnitude.
                                                               the direction in which it was moving at the
                •    Now, let the stone go by releasing the    time it was released, just like the piece of
                     thread.                                   stone in the activity described above. There
                •    Can you tell the direction in which       are many more familiar examples of objects
                     the stone moves after it is released?
                                                               moving under uniform circular motion,
                •    By repeating the activity for a few
                     times and releasing the stone at          such as the motion of the moon and the
                     different positions of the circular       earth, a satellite in a circular orbit around
                     path, check whether the direction in      the earth, a cyclist on a circular track at
                     which the stone moves remains the         constant speed and so on.
                     same or not.

                                    What

                                    you have

                                    learnt


                                    •     Motion is a change of position; it can be described in terms
                                          of the distance moved or the displacement.
                                    •     The motion of an object could be uniform or non-uniform
                                          depending on whether its velocity is constant or changing.

                                    •     The speed of an object is the distance covered per unit time,
                                          and velocity is the displacement per unit time.
                                    •     The acceleration of an object is the change in velocity per
                                          unit time.
                                    •     Uniform and non-uniform motions of objects can be shown
                                          through graphs.
                                    •     The motion of an object moving at uniform acceleration can
                                          be described with the help of the following equations, namely
                                                  v = u + at
                                                  s = ut + ½ at 2
                                                          2
                                                  2as = v – u 2
                  84                                                                                  SCIENCE





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