Question

the position vector of a moving particle at 't' sec is given by rbar =3icap+4t^2jcap-t^3kcap.its displacement during an interval of t= one sto 3sec is..
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Answer 1

ANSWER:

• $Displacement=32\hat{j}-26\hat{k}$

GIVEN:

• $\vec{r} =3\hat{i} +4t^2 \hat{j}-t^3 \hat{k}$.

• Interval t = 1 s to 3 s.

TO FIND:

• Displacement of the particle.

EXPLANATION:

Displacement = Final position - initial position

${Displacement = \vec{r} \ at \ 3 \: s - \vec{r} \ at \ 1 \: s}$

$\vec{r} \ at \ 3 \: s = 3 \hat{i} + 4( {3}^{2} ) \hat{j} - ( { 3}^{3} ) \hat{k}$

$\vec{r} \ at \ 3 \: s = 3 \hat{i} + 4( 9 ) \hat{j} -27 \hat{k}$

$\vec{r} \ at \ 3 \: s = 3 \hat{i} + 36 \hat{j} -27 \hat{k}$

$\vec{r} \ at \ 1 \: s = 3 \hat{i} + 4( {1}^{2} ) \hat{j} - ( {1}^{3} ) \hat{k}$

$\vec{r} \ at \ 1 \: s = 3 \hat{i} + 4 \hat{j} - \hat{k}$

$\vec{r} \ at \ 3 \: s - \vec{r} \ at \ 1 \: s = d$

${d = (3 - 3) \hat{i} + (36 - 4) \hat{j} - (27 - 1) \hat{k}}$

$d=0\hat{i}+32\hat{j}-26\hat{k}$

$Displacement=32\hat{j}-26\hat{k}$

$\large{\bold{\therefore \ Displacement=32\hat{j}-26\hat{k}}}$