The time taken for the velocity vectors of two bodies to become perpendicular to each other is:

0.1 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.2 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.4 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.8 s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $0.8 s$
Velocity of first body at any instant $t, {\to v}_{1} = 2^i - g t^j$
Velocity of second body at any instant
${\to v}_{2} = - 8^i - g t^j$
Since, ${\to v}_{1} ⊥ {\to v}_{2}$ , i.e., ${\to v}_{1} \cdot {\to v}_{2} = 0$
$(2^i - g t^j) \cdot (- 8^i - g t^j) = 0 \Rightarrow - 16 + g^{2} t^{2} = 0$
$\Rightarrow t = \sqrt{\frac{16}{g^{2}}} \Rightarrow t = \frac{4}{10} = 0.4 s$
${\to S}_{1} = (2 \times 0.4)^i, {\to S}_{2} = (8 \times 0.4) (-^i)$
${\to S}_{1} = 0.8^i$ and ${\to S}_{2} = - 3.2^i$
Separation $= {\to S}_{1} - (- {\to S}_{2}) = 0.8^i + (3.2^i) = 4^i$
${\to S}_{1} = 2 t^i - \frac{1}{2} g t^{2}^i, {\to S}_{2} = - 8 t^i - \frac{1}{2} g t^{2}^j$
As ${\to S}_{1} ⊥ {\to S}_{2}$ i.e., ${\to S}_{1} \cdot {\to S}_{2} = 0$
$- 16 t + \frac{1}{2} \times \frac{1}{2} g^{2} t^{4} = 0 \Rightarrow g^{2} t^{2} = 16 \times 4$
$\Rightarrow g t = 4 \times 2 \Rightarrow t = \frac{8}{10} = 0.8 s$ .

Suggest Corrections

Similar questions

\to r = (t^{2} + 4 t + 6)^i + (t^{2})^j .

{(\frac{0.4 \times 0.4 \times 0.4 + 0.2 \times 0.2 \times 0.2}{0.2 \times 0.2 \times 0.2 + 0.1 \times 0.1 \times 0.1})}^{3}

0.8 M

0.4 M

15

0.1 M

0.025 M

Join BYJU'S Learning Program