Consider the following velocity and time graph: Which one of the following is the value of average acceleration from 8 s to 12 s?
-1 m/s 2
Concept-
Velocity:
- The rate of change of displacement of a body is called as velocity of that body.
- It is a vector quantity which has both magnitude as well as direction.
\({\rm{Velocity\;}}\left( {\rm{V}} \right) = \frac{{Change\;in\;displacement}}{{Time\;taken}} = \frac{{{S_2} - {S_1}}}{{{t_2} - {t_1}}}\)
Where S 2is final displacement at time t 2and S 1is initial displacement at time t 1.
Acceleration: The rate of change of velocity is called as acceleration of the body.
\(a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\)
Where v 2= Final velocity and v 1= Initial velocity of an object at time respectively
Average acceleration:
The net rate of change in velocity over a period of time is called as average acceleration of the body.
\({a_{avg}} = \frac{{Final\;velocity\left( {{V_2}} \right) - Inital\;velocity\left( {{V_1}} \right)}}{{final\;time\left( {{t_2}} \right) - initial\;time\left( {{t_1}} \right)}}\)
Explanation-
From the graph we can see that:
Velocity at time 8s = initial velocity = V 1= 8 m/s and initial time t 1= 8s
Velocity time 12s = final velocity = V 2= 4m/s and final time t 2= 12s
Average acceleration,
\({a_{avg}} = \frac{{Final\;velocity\left( {{V_2}} \right) - Inital\;velocity\left( {{V_1}} \right)}}{{final\;time\left( {{t_2}} \right) - initial\;time\left( {{t_1}} \right)}} = \frac{{4 - 8}}{{12 - 8}}\)
\({a_{avg}} = - 1\;m/s\)
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