Published by:
CGP EDU Academic Team
Published on: September 11, 2026
A particle is moving along the x-axis with its coordinate with time t given by x(t) = 10 + 8t – 3t 2 . Another particle is moving along the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t 3 .
At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as 
. Then v (in m/s) is
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: Find the velocity of the first particle. \( x(t) = 10 + 8t - 3t^2 \)
Differentiate to find the velocity: \( v_x(t) = \frac{dx}{dt} = 8 - 6t \)
At \( t = 1 \): \( v_x(1) = 8 - 6(1) = 2 \, \text{m/s} \)
Step 2: Find the velocity of the second particle. \( y(t) = 5 - 8t^3 \)
Differentiate to find the velocity: \( v_y(t) = \frac{dy}{dt} = -24t^2 \)
At \( t = 1 \): \( v_y(1) = -24(1)^2 = -24 \, \text{m/s} \)
Step 3: Speed of the second particle in the frame of the first particle is given by:
\( v' = v_y + v_x \)
Substitute values: \( v' = -24 + 2 = -22 \, \text{m/s} \)
The magnitude is \( |v'| = 22 \, \text{m/s} \)
Therefore, \( v = 22 \, \text{m/s} \).
Therefore, A.
Differentiate to find the velocity: \( v_x(t) = \frac{dx}{dt} = 8 - 6t \)
At \( t = 1 \): \( v_x(1) = 8 - 6(1) = 2 \, \text{m/s} \)
Step 2: Find the velocity of the second particle. \( y(t) = 5 - 8t^3 \)
Differentiate to find the velocity: \( v_y(t) = \frac{dy}{dt} = -24t^2 \)
At \( t = 1 \): \( v_y(1) = -24(1)^2 = -24 \, \text{m/s} \)
Step 3: Speed of the second particle in the frame of the first particle is given by:
\( v' = v_y + v_x \)
Substitute values: \( v' = -24 + 2 = -22 \, \text{m/s} \)
The magnitude is \( |v'| = 22 \, \text{m/s} \)
Therefore, \( v = 22 \, \text{m/s} \).
Therefore, A.
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