Question

When the motion of a particle in x-y plane is defined by r = t²î + (4t-t³)ĵ where t is in sec and r is in meter.
Find (a) Average velocity in the time interval 1 sce to 3 sec. (b) Velocity of the particle at t= 3 sec. (c) Average acceleration in the time interval 1 sec to 3 sec. (d) Acceleration the particle at t = 3 sec.​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

The position of the particle is defined by the function:

$\sf{r = {t}^{2} \hat{i} \: + (4t - {t}^{3}) \hat{j} } \\ \\ \rightarrow \: \sf{r = {t}^{2} \hat{i} + 4t \hat{j} - {t}^{3} \hat{j} }$

To find

• Average Velocity

• Average Acceleration

Differentiating r w.r.t to t,we get velocity of the particle :

$\sf{ \vec{v} = \frac{d(\sf{{t}^{2} \hat{i} + 4t \hat{j} - {t}^{3} \hat{j} })}{dt} } \\ \\ \rightarrow \ \boxed{\sf{ \vec{v} = 2t \hat{i} \: + 4 \hat{j} - 3t {}^{2} \hat{j} }}$

• When t = 1s,

$\sf{ \vec{v} = 2(1) \hat{i} + 4 \hat{j} - 3(1) \hat{j}} \\ \\ \implies \: \sf{ \vec{v} =( 2 \hat{i} + \hat{j}) \: {ms}^{ - 1} }$

• When t = 3s,

$\sf{ \vec{v} = 2(3) \hat{i} + 4 \hat{j} - 3(3) \hat{j}} \\ \\ \implies \: \sf{ \vec{v} = (6 \hat{i} - 5 \hat{j}) \: ms {}^{ - 1} }$

Thus,the velocity vector at t = 3s is (6î - 5j) m/s

Average Velocity

$\displaystyle{\sf{ \vec{ \bar{v}} = \frac{(6 \hat{i} - 5 \hat{j}) - (2 \hat{i} + \hat{j})}{3 - 1} } }\\ \\ \displaystyle{ \leadsto \: \sf{ \vec{ \bar{v}} \: = \frac{4 \hat{i} - 6 \hat{j}}{2} }} \\ \\ \huge{ \leadsto \: \green{\sf{ \vec{ \bar{v}}} = (2 \hat{i} - 3 \hat{j}) \: ms {}^{ - 1}}}$

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Differentiating v w.r.t to t,we get acceleration of the particle:

$\sf{ \vec{a} = \frac{d(2t \hat{i} \: + 4 \hat{j} - 3t {}^{2} \hat{j} )}{dt} } \\ \\ \rightarrow \: \boxed{ \sf{ \vec{a} =( 2 \hat{i} - 6t \hat{j}) \: ms {}^{ - 2} }}$

• When t = 1s,

$\sf{ \vec{a} =( 2 \hat{i} - 6 \: \hat{j}) \: {ms}^{ - 2} }$

• When t = 3s,

$\sf{ \vec{a} = 2 \hat{i} - 12 \hat{j}}$

Acceleration vector of the particle at t = 3s is (2î - 12j)m/s²

Average Acceleration

$\sf{ \vec{ \bar{a}} = \frac{(2 \hat{i} - 12 \hat{j}) - (2 \hat{i} - 6 \hat{j})}{3 - 1} } \\ \\ \huge{\leadsto \: \sf{ \green{ \vec{\bar{a}} \: = - 3 \hat{j} \: ms {}^{ - 2} }}}$

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