Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1073: NEET 2016 Physics - Class 11 NEET-2016-Ph-1073 Solution: If the velocity of a particle is \(v = At + Bt^2\) where A and…

If the velocity of a particle is v=At+Bt2v = At + Bt^2 where A and B are constants then the distance travelled by it between 1 s and 2 s is 32A+73B\frac{3}{2}A + \frac{7}{3}B A2+B3\frac{A}{2} + \frac{B}{3} 32A+4B\frac{3}{2}A + 4B 3A+7B3A + 7B

Multiple Choice Options

  • A 32A+73B{3 \over 2}A + {7 \over 3}B 2 3 _ A + 3 7 _ B
  • B A2+B3{A \over 2} + {B \over 3} 2 A _ + 3 B _
  • C 32A+4B{3 \over 2}A + 4B 2 3 _ A + 4 B
  • D 3A+7B3A + 7B 3 A + 7 B

Step-by-Step Verified Solution

Given, v = At + Bt 2 d x d t = At + Bt 2 � d x = � ( A t + B t 2 ) d t x = A t 2 2 + B t 3 3 + C At t = 1, particle is at x(t = 1) = A 2 + B 3 + C At t = 2, particle is at x(t = 2) = 2 A + 8 B 3 + C _ distance travelled by the particle between 1 s and 2 s is, = x(t = 2) - x(t = 1) = ( 2 A + 8 B 3 + C ) - ( A 2 + B 3 + C ) = 3 2 A + 7 3 B

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