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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minQB

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Passage / Comprehension

A student designed a rocket that was launched straight up into the air from the ground. At $t = 0$ , the rocket is at $y = 0$ with $V_{y} (t = 0) = 0$ . The velocity of the rocket is given by: $V_{y} = (24 t - 3 t^{2}) m / s$ for $0 \leq t \leq t_{b}$ , where $t_{b}$ is the time at which the fuel burns out. The vertically upward direction is taken as positive. (Assume $g = 10 m / s^{2}$ )

The displacement of the rocket till the fuel burns out $(t = t_{b})$ is

Options:

Answer:

Solution:

At $t = t_{b}, a = - g = - 10$ So $24 - 6 t_{b} = - 10$ or $t_{b} = \frac{17}{3} sec$ $y = \int v_{y} dt = \int_{0}^{17/3} (24 t - 3 t^{2}) \cdot dt = (12 t^{2} - t^{3}) \int_{0}^{17/3}$

$= 12 ((\frac{17}{3})^{2} - 0^{2}) - ((\frac{17}{3})^{3} - 0^{3}) \approx 203.37$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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