Question:
A particle is moving along the x-axis with its coordinate with the time 't' given be \(x(t)=10+8t-3t^2\). Another particle is moving the y-axis with its coordinate as a function of time given by \(y(t)=5-8t^3\).
At \(t=1\) s, 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 ______.
A particle is moving along the x-axis with its coordinate with the time 't' given be \(x(t)=10+8t-3t^2\). Another particle is moving the y-axis with its coordinate as a function of time given by \(y(t)=5-8t^3\).
At \(t=1\) s, 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 ______.
At \(t=1\) s, 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 ______.
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Find both velocities at t = 1 s, subtract (second minus first as vectors), and square the magnitude of the relative velocity to get \(v\).
Updated On: Nov 5, 2026
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Correct Answer: 580
Solution and Explanation
Velocity of first particle: \(v_x=\dfrac{dx}{dt}=8-6t\Rightarrow v_x(1)=2\) m/s along +x.
Velocity of second particle: \(v_y=\dfrac{dy}{dt}=-24t^2\Rightarrow v_y(1)=-24\) m/s along y.
Velocity of 2nd relative to 1st: \(\vec v_{21}=\vec v_2-\vec v_1=(-2)\hat i+(-24)\hat j\).
Speed \(=\sqrt{(-2)^2+(-24)^2}=\sqrt{4+576}=\sqrt{580}\).
Comparing with \(\sqrt{v}\): \(v=580\).
Answer: 580
Velocity of second particle: \(v_y=\dfrac{dy}{dt}=-24t^2\Rightarrow v_y(1)=-24\) m/s along y.
Velocity of 2nd relative to 1st: \(\vec v_{21}=\vec v_2-\vec v_1=(-2)\hat i+(-24)\hat j\).
Speed \(=\sqrt{(-2)^2+(-24)^2}=\sqrt{4+576}=\sqrt{580}\).
Comparing with \(\sqrt{v}\): \(v=580\).
Answer: 580
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