The position of a particle related to time is given by x = (5t² - 4t + 5) m. The magnitude of velocity of the particle at t = 2 s will be: (a) 14 ms⁻¹ (b) 16 ms⁻¹ (c) 10 ms⁻¹ (d) 06 ms⁻¹

Answer: (b). v = dx/dt = 10t - 4 At t = 2 s: v = 20 - 4 = 16 m/s So the answer is (b) 16 ms⁻¹.

The position of a particle related to time is given by x = (5t² - 4t… | JEE Main 2023 (15 Apr, Shift 1)

Q18JEE Main 2023 · 15 APR · SHIFT 1Easy

Velocity, Acceleration & Variable Motion

The position of a particle related to time is given by $x = \left(5t^2 - 4t + 5\right)\text{ m}$ . The magnitude of velocity of the particle at $t = 2\text{ s}$ will be:

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Answer: (b)

$v = \frac{dx}{dt} = 10t - 4$
At $t = 2$ s: $v = 20 - 4 = 16$ m/s
So the answer is (b) $16\text{ ms}^{-1}$ .

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