Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The acceleration of a particle is increasing linearly with time $t$ as $bt$ . The particle starts from the origin with an initial velocity of $v_0$ . The distance travelled by the particle in time $t$ will be:

$v_0t + \frac{1}{3}bt^2$

$v_0t + \frac{1}{3}bt^3$

$v_0t + \frac{1}{6}bt^3$

$v_0t + \frac{1}{2}bt^2$

Step-by-Step Solution

Identify the Relationship: Acceleration $a$ is given as a function of time: $a = bt$ . Since acceleration is non-uniform, we must use calculus (integration) to find velocity and position .
Find Velocity ( $v$ ): Acceleration is the derivative of velocity ( $a = \frac{dv}{dt}$ ). Integrating acceleration with respect to time: $\int_{v_0}^{v} dv = \int_{0}^{t} a \, dt = \int_{0}^{t} bt \, dt$ $v - v_0 = \left[ \frac{bt^2}{2} \right]_0^t \implies v = v_0 + \frac{1}{2}bt^2$
Find Distance ( $x$ ): Velocity is the derivative of position ( $v = \frac{dx}{dt}$ ). Integrating velocity with respect to time: $\int_{0}^{x} dx = \int_{0}^{t} v \, dt = \int_{0}^{t} \left( v_0 + \frac{1}{2}bt^2 \right) dt$ $x = \left[ v_0t + \frac{1}{2}b \cdot \frac{t^3}{3} \right]_0^t$ $x = v_0t + \frac{1}{6}bt^3$

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