cbse class 9 science exploration chapter 6 solutions – Discussion Forum

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Sonal Chauhan

Asked: September 16, 2026In: Class 9 Science

The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.

From Fig. 6.41, the slope of the velocity-time line gives an acceleration of 2.5 m s-2. Multiplying this acceleration by the object mass of 10 kg gives a constant acting force of 25 N.

Sonal Chauhan

Asked: September 16, 2026In: Class 9 Science

When a net force acts on an object, we observe that the object accelerates: (i) opposite to the direction of force, with acceleration proportional to the force acting on the object. (ii) opposite to the direction of force, with acceleration proportional to the mass of the object. (iii) in the direction of force, with acceleration inversely proportional to the force acting on the object. (iv) in the direction of force, with acceleration proportional to the force acting on the object.

Option (iv) is correct because Newton’s second law of motion establishes that when a net force acts on an object, it produces acceleration in the direction of the force directly proportional to its magnitude.