Question

In an experiment, students used video analysis to track the motion of an object falling vertically through a fluid in a glass cylinder. The object of m equals 12 grams is released from rest at the top of the column of fluid, as shown above. The data for the speed v of the falling object as a function of time t are graphed on the grid below. The dashed curve represents the best fit chosen by the students for these data 1.0 0.8 0.6 v (m/s) 0.4 0.2 0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 Using the graph, calculate an approximate value for the magnitude of the acceleration of the object at t= 0.20 s and at t=0.05s.

Answer 1

Acceleration is 2.22 m/$s^{2}$ at t =0.20 s. We can use the graph to know the value of the acceleration.

The time rate of change of velocity is known as acceleration. An object accelerated when its speed is increasing or decreasing or it is changing direction of motion. The car is moving forward in a positive direction and speeding up is an example of acceleration.

acceleration = change in velocity/ change in time

a = ΔV/ΔT

So, in our question,

a = 0.8 - 0.6/0,2-0.3

   = 2.22 m$s^{2}$

There are three types of accelerated motions : 1) uniform acceleration, 2)non-uniform acceleration and 3)average acceleration.

Acceleration is a vector quantity. Its SI unit is m/$s^{2}$

To know more about acceleration here

brainly.com/question/2303856

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