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If the velocity of a particle is v=At+Bt2v = At + Bt^2, where AA and BB are constants, then the distance travelled by it between 1 s and 2 s is:

A

3A + 7B

B

3A/2 + 7B/3

C

A/2 + B/3

D

3A/2 + 4B

Step-by-Step Solution

Velocity is defined as the rate of change of position with respect to time (v=dxdtv = \frac{dx}{dt}) . To find the distance travelled between two instants, we integrate the velocity function over the given time interval.

  1. Set up the integral: Distance=t1t2vdt=12(At+Bt2)dt\text{Distance} = \int_{t_1}^{t_2} v \, dt = \int_{1}^{2} (At + Bt^2) \, dt

  2. Integrate: Using the power rule tndt=tn+1n+1\int t^n dt = \frac{t^{n+1}}{n+1}: x=[At22+Bt33]12x = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}

  3. Apply limits (Upper limit 2, Lower limit 1): x=(A(2)22+B(2)33)(A(1)22+B(1)33)x = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right) x=(4A2+8B3)(A2+B3)x = \left( \frac{4A}{2} + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right) x=2A+8B3A2B3x = 2A + \frac{8B}{3} - \frac{A}{2} - \frac{B}{3}

  4. Simplify: x=(2AA2)+(8B3B3)x = \left( 2A - \frac{A}{2} \right) + \left( \frac{8B}{3} - \frac{B}{3} \right) x=3A2+7B3x = \frac{3A}{2} + \frac{7B}{3}

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