The position vector of a moving body at any instant of time is given as . The magnitude and direction of velocity at t = 2 s is, [2025]

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Q.
The position vector of a moving body at any instant of time is given as $\vec{r} = (5 t^{2} \hat{i} - 5 t \hat{j}) m$ . The magnitude and direction of velocity at t = 2 s is, [2025]

1 $5 \sqrt{15}$ m/s, making an angle of $\tan^{- 1}$ 4 with –ve Y axis

2 $5 \sqrt{15}$ m/s, making an angle of $\tan^{- 1}$ 4 with +ve X axis

3 $5 \sqrt{17}$ m/s, making an angle of $\tan^{- 1}$ 4 with –ve Y axis

4 $5 \sqrt{17}$ m/s, making an angle of $\tan^{- 1}$ 4 with +ve X axis

Ans.
(3)

$\vec{r} = 5 t^{2} \hat{i} - 5 \hat{j}$

$\vec{v} = 10 t \hat{i} - 5 \hat{j}$

$\vec{v} = 20 \hat{i} - 5 \hat{j} at t = 2 \sec$

$| \vec{v} | = 5 \sqrt{17} m / s$

$\tan θ = \frac{20}{5} = 4$

$θ = \tan^{- 1}$ 4 with –ve Y-axis

Solution

Q.
The position vector of a moving body at any instant of time is given as $\vec{r} = (5 t^{2} \hat{i} - 5 t \hat{j}) m$ . The magnitude and direction of velocity at t = 2 s is, [2025]

1 $5 \sqrt{15}$ m/s, making an angle of $\tan^{- 1}$ 4 with –ve Y axis

2 $5 \sqrt{15}$ m/s, making an angle of $\tan^{- 1}$ 4 with +ve X axis

3 $5 \sqrt{17}$ m/s, making an angle of $\tan^{- 1}$ 4 with –ve Y axis

4 $5 \sqrt{17}$ m/s, making an angle of $\tan^{- 1}$ 4 with +ve X axis

Ans.
(3)

$\vec{r} = 5 t^{2} \hat{i} - 5 \hat{j}$

$\vec{v} = 10 t \hat{i} - 5 \hat{j}$

$\vec{v} = 20 \hat{i} - 5 \hat{j} at t = 2 \sec$

$| \vec{v} | = 5 \sqrt{17} m / s$

$\tan θ = \frac{20}{5} = 4$

$θ = \tan^{- 1}$ 4 with –ve Y-axis

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Solution

Q. The position vector of a moving body at any instant of time is given as r→=(5t2i^–5tj^) m. The magnitude and direction of velocity at t = 2 s is, [2025]

1 515 m/s, making an angle of tan–1 4 with –ve Y axis

2 515 m/s, making an angle of tan–1 4 with +ve X axis

3 517 m/s, making an angle of tan–1 4 with –ve Y axis

4 517 m/s, making an angle of tan–1 4 with +ve X axis

Ans. (3) r→=5t2i^–5j^ v→=10ti^–5j^ v→=20i^–5j^ at t=2 sec |v→|=517 m/s tanθ=205=4 θ=tan–1 4 with –ve Y-axis

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Q.
The position vector of a moving body at any instant of time is given as $\vec{r} = (5 t^{2} \hat{i} - 5 t \hat{j}) m$ . The magnitude and direction of velocity at t = 2 s is, [2025]

1 $5 \sqrt{15}$ m/s, making an angle of $\tan^{- 1}$ 4 with –ve Y axis

2 $5 \sqrt{15}$ m/s, making an angle of $\tan^{- 1}$ 4 with +ve X axis

3 $5 \sqrt{17}$ m/s, making an angle of $\tan^{- 1}$ 4 with –ve Y axis

4 $5 \sqrt{17}$ m/s, making an angle of $\tan^{- 1}$ 4 with +ve X axis

Ans.
(3)

$\vec{r} = 5 t^{2} \hat{i} - 5 \hat{j}$

$\vec{v} = 10 t \hat{i} - 5 \hat{j}$

$\vec{v} = 20 \hat{i} - 5 \hat{j} at t = 2 \sec$

$| \vec{v} | = 5 \sqrt{17} m / s$

$\tan θ = \frac{20}{5} = 4$

$θ = \tan^{- 1}$ 4 with –ve Y-axis