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Let r(1)(t)=3t hat(i)+4t^(2)hat(j) and...

Let `r_(1)(t)=3t hat(i)+4t^(2)hat(j)`
and `r_(2)(t)=4t^(2) hat(i)+3t^()hat(j)`
represent the positions of particles 1 and 2, respectively, as function of time t, `r_(1)(t)` and `r_(2)(t)` are in meter and t in second. The relative speed of the two particle at the instant t = 1s, will be

A

1 m/s

B

`3sqrt(2) m//s`

C

`5sqrt(2) m//s`

D

`7sqrt(2)m//s`

Text Solution

AI Generated Solution

The correct Answer is:
To find the relative speed of the two particles at the instant \( t = 1 \) second, we will follow these steps: ### Step 1: Determine the position functions The position functions of the two particles are given as: - \( \mathbf{r}_1(t) = 3t \hat{i} + 4t^2 \hat{j} \) - \( \mathbf{r}_2(t) = 4t^2 \hat{i} + 3t \hat{j} \) ### Step 2: Differentiate the position functions to find velocity To find the velocities of the two particles, we differentiate their position functions with respect to time \( t \). For particle 1: \[ \mathbf{v}_1(t) = \frac{d\mathbf{r}_1}{dt} = \frac{d}{dt}(3t \hat{i} + 4t^2 \hat{j}) = 3 \hat{i} + 8t \hat{j} \] For particle 2: \[ \mathbf{v}_2(t) = \frac{d\mathbf{r}_2}{dt} = \frac{d}{dt}(4t^2 \hat{i} + 3t \hat{j}) = 8t \hat{i} + 3 \hat{j} \] ### Step 3: Evaluate the velocities at \( t = 1 \) second Now, we substitute \( t = 1 \) into the velocity equations. For particle 1: \[ \mathbf{v}_1(1) = 3 \hat{i} + 8(1) \hat{j} = 3 \hat{i} + 8 \hat{j} \] For particle 2: \[ \mathbf{v}_2(1) = 8(1) \hat{i} + 3 \hat{j} = 8 \hat{i} + 3 \hat{j} \] ### Step 4: Calculate the relative velocity The relative velocity \( \mathbf{v}_{\text{relative}} \) is given by: \[ \mathbf{v}_{\text{relative}} = \mathbf{v}_1 - \mathbf{v}_2 \] Substituting the values we found: \[ \mathbf{v}_{\text{relative}} = (3 \hat{i} + 8 \hat{j}) - (8 \hat{i} + 3 \hat{j}) = (3 - 8) \hat{i} + (8 - 3) \hat{j} = -5 \hat{i} + 5 \hat{j} \] ### Step 5: Find the magnitude of the relative velocity To find the relative speed, we calculate the magnitude of the relative velocity vector: \[ |\mathbf{v}_{\text{relative}}| = \sqrt{(-5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \] ### Conclusion The relative speed of the two particles at the instant \( t = 1 \) second is \( 5\sqrt{2} \) meters per second. ---

To find the relative speed of the two particles at the instant \( t = 1 \) second, we will follow these steps: ### Step 1: Determine the position functions The position functions of the two particles are given as: - \( \mathbf{r}_1(t) = 3t \hat{i} + 4t^2 \hat{j} \) - \( \mathbf{r}_2(t) = 4t^2 \hat{i} + 3t \hat{j} \) ### Step 2: Differentiate the position functions to find velocity ...
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