Question 156: Motion in 2D & Vectors

Question

A particle is moving along the x-axis with its
coordinate with the time 't' given be x(t) = 10 + 8t – 3t2. Another particle is moving
the y-axis with its coordinate as a function of
time given by y(t) = 5 – 8t3. At t = 1s, the speed
of the second particle as measured in the frame
of the first particle is given as $$\sqrt v $$. Then v
(in m/s) is ______.

Accepted Answer

For a particle Ã¢ÂÂAÃ¢ÂÂ, its position along the x-axis as a function of time $ t $ is given by:
$ x(t) = 10 + 8t - 3t^2 $
To find the velocity $ v_A $, we take the derivative of $ x(t) $ with respect to $ t $:
$ v_A = \frac{d}{dt}[10 + 8t - 3t^2] = 8 - 6t $
At $ t = 1 $ second, the velocity of particle A is:
$ \vec{v_A} = (8 - 6 \cdot 1)\hat{i} = 2\hat{i} $
For a particle Ã¢ÂÂBÃ¢ÂÂ, its position along the y-axis as a function of time $ t $ is given by:
$ y(t) = 5 - 8t^3 $
To find the velocity $ v_B $, we take the derivative of $ y(t) $ with respect to $ t $:
$ v_B = \frac{d}{dt}[5 - 8t^3] = -24t^2 $
At $ t = 1 $ second, the velocity of particle B is:
$ \vec{v_B} = -24 \cdot 1^2 \hat{j} = -24\hat{j} $
The velocity of particle B relative to particle A ($ \vec{v_{B/A}} $) is calculated as:
$ \vec{v_{B/A}} = \vec{v_B} - \vec{v_A} $
$ \vec{v_{B/A}} = -24\hat{j} - 2\hat{i} $
To find the magnitude of $ \vec{v_{B/A}} $:
$ |\vec{v_{B/A}}| = \sqrt{(-24)^2 + (-2)^2} = \sqrt{576 + 4} = \sqrt{580} $
Therefore, $ v $ is:
$ v = 580 $

Motion in 2D & Vectors

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