
A particle is moving such that its position coordinates (x,y) are (2m, 3m) at time t=0, (6m, 7m) at time t=2 s and (13m, 14m) at time t=5 s. Average velocity vector (${\vec{V}}_{avg}$) from t=0 to t=5 s is:
A. $\dfrac {1}{5}(13\hat{i} + 14\hat{j})$
B. $\dfrac {7}{3}(\hat{i} + \hat{j})$
C. $2(\hat{i} - \hat{j})$
D. $\dfrac {11}{5}(\hat{i} + \hat{j})$
Answer
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Hint: To solve this problem, first find the net displacement of the particle. To find the net, subtract the initial position vector of the particle from the final position vector of the particle. Then, find the time taken for this net displacement by the particle. Average velocity is the ratio of net displacement to the time taken. So, substitute the values and find the average velocity of the particle.
Formula used:
${\vec{V}}_{avg}= \dfrac {\Delta r}{T}$
Complete step-by-step solution:
At time t=0, position vector of the particle is given by,
${r}_{1}= 2\hat{i} + 3\hat{j}$
At time t=5, position vector of the particle is given by,
${r}_{2}= 13\hat{i} + 14\hat{j}$
Net displacement is given by,
$\Delta r= {r}_{2}-{r}_{1}$
Substituting values in above equation we get,
$\Delta r= 13\hat{i} + 14\hat{j} -2\hat{i} + 3\hat{j}$
$\Rightarrow \Delta r= (13-2)\hat {i} +(14-3)\hat{j}$
$\Rightarrow \Delta r= 11\hat {I}+ 11 \hat{j}$
Time taken for the displacement,
$T= 5-0$
$\Rightarrow T=5 s$
Average velocity is given by,
${\vec{V}}_{avg}= \dfrac {\Delta r}{T}$
Substituting the values in above equation we get,
${\vec{V}}_{avg}= \dfrac {11\hat {i}+ 11 \hat{j}}{5}$
$\Rightarrow {\vec{V}}_{avg}= \dfrac {11}{5}(\hat{i} + \hat{j})$
Thus, the average velocity vector (${\vec{V}}_{avg}$) from t=0 to t=5 s is $\dfrac {11}{5}(\hat{i} + \hat{j})$.
So, the correct answer is option D i.e. $\dfrac {11}{5}(\hat{i} + \hat{j})$.
Note: Students must not confuse between average velocity and velocity of a particle. The average velocity of a particle is defined as the net displacement of a particle or a body in a certain time interval. Whereas, velocity is defined as the displacement at that instant of time. As both are a ratio of displacement to time, their units are the same.
Formula used:
${\vec{V}}_{avg}= \dfrac {\Delta r}{T}$
Complete step-by-step solution:
At time t=0, position vector of the particle is given by,
${r}_{1}= 2\hat{i} + 3\hat{j}$
At time t=5, position vector of the particle is given by,
${r}_{2}= 13\hat{i} + 14\hat{j}$
Net displacement is given by,
$\Delta r= {r}_{2}-{r}_{1}$
Substituting values in above equation we get,
$\Delta r= 13\hat{i} + 14\hat{j} -2\hat{i} + 3\hat{j}$
$\Rightarrow \Delta r= (13-2)\hat {i} +(14-3)\hat{j}$
$\Rightarrow \Delta r= 11\hat {I}+ 11 \hat{j}$
Time taken for the displacement,
$T= 5-0$
$\Rightarrow T=5 s$
Average velocity is given by,
${\vec{V}}_{avg}= \dfrac {\Delta r}{T}$
Substituting the values in above equation we get,
${\vec{V}}_{avg}= \dfrac {11\hat {i}+ 11 \hat{j}}{5}$
$\Rightarrow {\vec{V}}_{avg}= \dfrac {11}{5}(\hat{i} + \hat{j})$
Thus, the average velocity vector (${\vec{V}}_{avg}$) from t=0 to t=5 s is $\dfrac {11}{5}(\hat{i} + \hat{j})$.
So, the correct answer is option D i.e. $\dfrac {11}{5}(\hat{i} + \hat{j})$.
Note: Students must not confuse between average velocity and velocity of a particle. The average velocity of a particle is defined as the net displacement of a particle or a body in a certain time interval. Whereas, velocity is defined as the displacement at that instant of time. As both are a ratio of displacement to time, their units are the same.
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