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# At time $t = {t_1}$ an object’s velocity is given by the vector $\overrightarrow {{v_1}}$ shown below.$\to$A short time later, at $t = {t_2}$, the object’s velocity is the vector $\overrightarrow {{v_2}}$$\nearrow$If $\overrightarrow {{v_1}}$ and $\overrightarrow {{v_2}}$ have the same magnitude, which one of the following vectors best illustrates the object’s average acceleration between $t = {t_1}$ and $t = {t_2}$?(A)(B)(C) (D) (E)

Last updated date: 22nd Jun 2024
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Hint: Since the magnitudes of the vectors are the same so we need to only consider their direction. In order to calculate the acceleration, the direction of the first vector must be reversed to represent the difference.

We are given the velocity vectors at time ${t_1}$ and ${t_2}$
So, $\overrightarrow a = \dfrac{{\overrightarrow {{v_2}} - \overrightarrow {{v_1}} }}{{{t_2} - {t_1}}}$
Now, the resultant of vectors $\overrightarrow {{v_1}}$ and $\overrightarrow {{v_2}}$results in option A. Since we need to find their difference, we can assume that the direction of vector $\overrightarrow {{v_1}}$ is reversed.