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Motion in A Straight Line question

2023 · 15 Apr · Shift 1 · Q46
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  5. /2023 · 15 Apr · Shift 1 · Q46

Motion in A Straight Line question

2023 · 15 Apr · Shift 1 · Q46

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
The position of a particle related to time is given by x=(5t2−4t+5)mx=\left(5 t^{2}-4 t+5\right) \mathrm{m}x=(5t2−4t+5)m. The magnitude of velocity of the particle at t=2st=2 st=2s will be :
  1. A
    14 ms−114 \mathrm{~ms}^{-1}14 ms−1
  2. B
    16 ms−116 \mathrm{~ms}^{-1}16 ms−1
  3. C
    10 ms−110 \mathrm{~ms}^{-1}10 ms−1
  4. D
    06 ms−106 \mathrm{~ms}^{-1}06 ms−1
View written solutionFree

Correct answer: B

  1. The position of the particle is given by x(t)=5t2−4t+5x(t)=5t^2-4t+5x(t)=5t2−4t+5

  2. Velocity is the time derivative of position: v(t)=dxdtv(t)=\frac{dx}{dt}v(t)=dtdx​

  3. Differentiate x(t)x(t)x(t) with respect to ttt: v(t)=ddt(5t2−4t+5)=10t−4v(t)=\frac{d}{dt}(5t^2-4t+5)=10t-4v(t)=dtd​(5t2−4t+5)=10t−4

  4. At t=2 st=2\,\text{s}t=2s, v(2)=10(2)−4=20−4=16 m s−1v(2)=10(2)-4=20-4=16\,\text{m s}^{-1}v(2)=10(2)−4=20−4=16m s−1

  5. Since the question asks for the magnitude of velocity, ∣v∣=16 m s−1|v|=16\,\text{m s}^{-1}∣v∣=16m s−1

  6. Comparing with the options:

    • A: 14 m s−114\,\text{m s}^{-1}14m s−1
    • B: 16 m s−116\,\text{m s}^{-1}16m s−1
    • C: 10 m s−110\,\text{m s}^{-1}10m s−1
    • D: 6 m s−16\,\text{m s}^{-1}6m s−1

    Hence, the correct option is B.

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