Answer
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Hint: Acceleration is defined as the rate of change of velocity with respect to time whereas average acceleration is defined as the rate of change of velocity for a particular interval of time. Here, acceleration is denoted by $a$ and average acceleration is denoted by ${a_v}$ .
Complete answer:
Average acceleration is defined as the rate of change of velocity of an object with respect to a particular interval of time, which is given as
${a_{average}} = \dfrac{{\Delta v}}{t}$
Where, ${a_{average}}$ is the average acceleration, $\Delta v$ is the rate of change of velocity, and $t$ is the time taken.
We can also write the above equation as,
$a = \dfrac{{{v_{final}} - {v_{initial}}}}{t}$
$ \Rightarrow \,{v_{final}} - {v_{initial}} = at$
$ \Rightarrow \,\Delta v = at$
Now, putting the value of $$\Delta v$$ in the first equation, we get
${a_{average}} = \dfrac{{at}}{t}$
$\therefore{a_{average}} = a$
Where, $a$ is the uniform acceleration.
Therefore, we can say that the average acceleration for a uniformly accelerated motion is equal to the uniform acceleration.
Hence, the option D is the correct option.
Additional information:
A uniformly accelerated motion refers to the motion that remains constant no matter if the time changes or not. Some common examples of uniformly accelerated motion are a ball rolling down a slope, a skydiver jumping out of the plane, a ball is thrown from a ladder, and a bicycle whose brakes should be engaged. These examples do not maintain absolute uniform acceleration because of the interference of gravity or friction.
Now, if the object is in uniform accelerated motion and it is moving along X-axis, then the motion is uniformly accelerated motion in the horizontal plane whereas if the object is moving along Y-axis then the motion is uniformly accelerated motion in the vertical plane.
Note:If the object is showing different velocities which are ${v_1}$ , ${v_2}$ , ${v_3}$ ………. ${v_n}$ for the time intervals ${t_1}$ , ${t_2}$ , ${t_3}$ ,………… ${t_n}$ .
Therefore, the average acceleration is given by,
${a_{average}} = \dfrac{{{v_1} + {v_2} + {v_3} + .......... + {v_n}}}{{{t_1} + {t_2} + {t_3} + ........... + {t_n}}}$
Complete answer:
Average acceleration is defined as the rate of change of velocity of an object with respect to a particular interval of time, which is given as
${a_{average}} = \dfrac{{\Delta v}}{t}$
Where, ${a_{average}}$ is the average acceleration, $\Delta v$ is the rate of change of velocity, and $t$ is the time taken.
We can also write the above equation as,
$a = \dfrac{{{v_{final}} - {v_{initial}}}}{t}$
$ \Rightarrow \,{v_{final}} - {v_{initial}} = at$
$ \Rightarrow \,\Delta v = at$
Now, putting the value of $$\Delta v$$ in the first equation, we get
${a_{average}} = \dfrac{{at}}{t}$
$\therefore{a_{average}} = a$
Where, $a$ is the uniform acceleration.
Therefore, we can say that the average acceleration for a uniformly accelerated motion is equal to the uniform acceleration.
Hence, the option D is the correct option.
Additional information:
A uniformly accelerated motion refers to the motion that remains constant no matter if the time changes or not. Some common examples of uniformly accelerated motion are a ball rolling down a slope, a skydiver jumping out of the plane, a ball is thrown from a ladder, and a bicycle whose brakes should be engaged. These examples do not maintain absolute uniform acceleration because of the interference of gravity or friction.
Now, if the object is in uniform accelerated motion and it is moving along X-axis, then the motion is uniformly accelerated motion in the horizontal plane whereas if the object is moving along Y-axis then the motion is uniformly accelerated motion in the vertical plane.
Note:If the object is showing different velocities which are ${v_1}$ , ${v_2}$ , ${v_3}$ ………. ${v_n}$ for the time intervals ${t_1}$ , ${t_2}$ , ${t_3}$ ,………… ${t_n}$ .
Therefore, the average acceleration is given by,
${a_{average}} = \dfrac{{{v_1} + {v_2} + {v_3} + .......... + {v_n}}}{{{t_1} + {t_2} + {t_3} + ........... + {t_n}}}$
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