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Motion in A Straight Line question

2016 · Q160
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Motion in A Straight Line question

2016 · Q160

NEETPhysicsMotion in A Straight LineMCQ+4 / −1
If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is
  1. A
    32A+73B{3 \over 2}A + {7 \over 3}B23​A+37​B
  2. B
    A2+B3{A \over 2} + {B \over 3}2A​+3B​
  3. C
    32A+4B{3 \over 2}A + 4B23​A+4B
  4. D
    3A+7B3A + 7B3A+7B
View written solutionFree

Correct answer: A

Given, v = At + Bt2

dxdt{{dx} \over {dt}}dtdx​ = At + Bt2

∫dx=∫(At+Bt2)dt\int {dx = \int {\left( {At + B{t^2}} \right)} } dt∫dx=∫(At+Bt2)dt

x = At22+Bt33+C{{A{t^2}} \over 2} + {{B{t^3}} \over 3} + C2At2​+3Bt3​+C

At t = 1, particle is at

x(t = 1) = A2+B3+C{A \over 2} + {B \over 3} + C2A​+3B​+C

At t = 2, particle is at

x(t = 2) = 2A+8B3+C2A + {{8B} \over 3} + C2A+38B​+C

∴\therefore∴ distance travelled by the particle between 1 s and 2 s is,

= x(t = 2) - x(t = 1)

= (2A+8B3+C2A + {{8B} \over 3} + C2A+38B​+C) - (A2+B3+C{A \over 2} + {B \over 3} + C2A​+3B​+C)

= 32A+73B{3 \over 2}A + {7 \over 3}B23​A+37​B

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