Acceleration Due to Gravity

A particle moving with variable velocity is said to possess acceleration. When a particle executes non-uniform motion, its velocity changes. Acceleration is defined as the rate of change of velocity. It is a vector quantity.

Acceleration Due to Gravity

Objects thrown vertically upwards move up to a certain distance and then fall back to the ground. This is due to the earth's gravitational force. Due to gravitational force, all objects are accelerated towards the Earth. This uniform acceleration towards the Earth, irrespective of the mass is known as acceleration due to gravity and is denoted by 'g'. Equations of Motion of Objects under the Influence of Gravity ( Neglecting air Resistance ) Here, since acceleration of a body moving vertically, either upward or downward, is due to gravity, ‘a’ is substituted by ‘g’ and the displacement ‘s’ is substituted by ‘h’. Thus, we have v = u+gt s = ut+½ gt² v² = u² + 2gh Where, v is the final velocity, u is initial velocity, g is acceleration due to gravity and t is time taken. ‘g’ is chosen to be positive if the body is moving towards the Earth, i.e., downward and g is negative if the body is moving in upwards direction.

NOTE , It is purely a matter of choice and convenience that we choose a particular direction as positive. In the above case, for example, the final result will remain the same even if we choose the upward direction as positive and downward as negative or vice—versa.

Equations of Motion of a Body Dropped from a Certain Height

A freely falling body is one which moves only under the influence of gravity ( i.e., no other force acts on it ) like air fraction. Let us consider the special case of a freely falling body which is dropped or released from rest for which the initial velocity, u = 0, and acceleration, a = +g. Taking u = 0, a = h and a = +g, we can write the equations of motion for a freely falling body, as 1. v = u+at v = gt 2. s = ut+½at² h = ½gt² 3. v² = u² = 2as v² = 2gh From h = ½gt²; t = √2h ⁄ √g, Where ‘t’ is the time of descent.

Unit of acceleration

SI unit of acceleration metre per second per second (m s^-2)  and cgs unit of acceleration cm s^-2. By definition, acceleration,  a = change in velocity/Time = v/t

Example A car moves along a straight path with variable velocity as shown in the figure. When the car is at position A, its velocity is 10 m s−1 and when it is at position B, its velocity is 20 m s−1. If the car takes 5 seconds of time to move from A to B, find the acceleration of the car.

Solution Initial position of the car at position A = u = 10 m s−1 Final velocity of the car at position B = v = 20 m s−1 The change in velocity of the car, Δv = v = 20 m s−1 −10m s−1 = 10 m s−1 The time taken for the car to move from A to B, Δt = 5 s ∴ Acceleration of the car, a = Δv/Δt = 10 m s−1 = 2 m s-2