OnlineCBT Logo OnlineCBT

Physics · MOTION IN A PLANE

हिन्दी

The position vector of a particle moving in the xy-plane is given by \( \vec{r}(t) = (4t^2 - 2t)\hat{i} + (3t^3 + 5)\hat{j} \) in SI units. The magnitude of the displacement vector of the particle between time \( t = 1\text{ s} \) and \( t = 2\text{ s} \) is:

  1. A
    \( \sqrt{425}\text{ m} \)
  2. B
    \( \sqrt{541}\text{ m} \)
  3. C
    15 m
  4. D
    31 m
View Answer & Solution

Correct answer · Option B

\( \sqrt{541}\text{ m} \)

Explanation

At \( t = 1\text{ s} \), \( \vec{r}_1 = 2\hat{i} + 8\hat{j} \). At \( t = 2\text{ s} \), \( \vec{r}_2 = 12\hat{i} + 29\hat{j} \). Displacement \( \Delta\vec{r} = 10\hat{i} + 21\hat{j} \). Magnitude \( |\Delta\vec{r}| = \sqrt{10^2 + 21^2} = \sqrt{541}\text{ m} \).
Address Contact

+91 7877547686

E-mail

onlinecbtportal@gmail.com

Helpline Number

+91 7877547686


Click To Download
Get it on Google Play

ऐप डाउनलोड करने
के लिए Google Play पर
उपलब्ध है

WhatsApp Chat