The position vector of a moving particle is given by (3i^+4t2j^−t3k^)m

(a)r→=(3i^+4t2j^−t3k^)m; when t=0 r=3i^[ i.e it starts from (3,0,0)](b) r→when t = 1s isr→1=(3i^+4j^−k^);r→ when t = 3 s is r→2=3i^+36j^−27k^ displacement during t = 1s to 3s is s→=r→2−r→1=32j^−26k^ (c)v→=drdt=8tj^−3t2k^component of velocity along x-axis at t = 2s is zero(d)a→=dv→dt=8j^−6tk^ acceleration is uniform along y - axis acceleration is non-uniform along z - axis

The position vector of a moving particle is given by (3i^+4t2j^−t3k^)m

(a)r→=(3i^+4t2j^−t3k^)m; when t=0 r=3i^[ i.e it starts from (3,0,0)](b) r→when t = 1s isr→1=(3i^+4j^−k^);r→ when t = 3 s is r→2=3i^+36j^−27k^ displacement during t = 1s to 3s is s→=r→2−r→1=32j^−26k^ (c)v→=drdt=8tj^−3t2k^component of velocity along x-axis at t = 2s is zero(d)a→=dv→dt=8j^−6tk^ acceleration is uniform along y - axis acceleration is non-uniform along z - axis

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The position vector of a moving particle is given by $(3 \hat{i} + 4 t^{2} \hat{j} - t^{3} \hat{k}) m$

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It starts from $(3, 0, 0)$

Acceleration of particle is along $(- Z)$ axis only

velocity component along x-axis at t = 2s is zero

Displacement from t = 1s to 3s is $(32 \hat{j} - 26 \hat{k}) m$

answer is A, B, C.

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Detailed Solution

$(a) \vec{r} = (3 \hat{i} + 4 t^{2} \hat{j} - t^{3} \hat{k}) m; w h e n t = 0$

$r = 3 \hat{i}$ [ i.e it starts from $(3, 0, 0)$ ]

(b) $\vec{r}$ when t = 1s is ${\vec{r}}_{1} = (3 \hat{i} + 4 \hat{j} - \hat{k})$ ; $\vec{r}$ when t = 3 s is ${\vec{r}}_{2} = 3 \hat{i} + 36 \hat{j} - 27 \hat{k}$ displacement during t = 1s to 3s is $\vec{s} = {\vec{r}}_{2} - {\vec{r}}_{1} = 32 \hat{j} - 26 \hat{k}$