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Motion in A Straight Line question

2023 · 25 Jan · Shift 2 · Q49
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  5. /2023 · 25 Jan · Shift 2 · Q49

Motion in A Straight Line question

2023 · 25 Jan · Shift 2 · Q49

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
The distance travelled by a particle is related to time t as x=4t2x=4\mathrm{t}^2x=4t2. The velocity of the particle at t=5s is :-
  1. A
    25 ms−1\mathrm{25~ms^{-1}}25 ms−1
  2. B
    20 ms−1\mathrm{20~ms^{-1}}20 ms−1
  3. C
    8 ms−1\mathrm{8~ms^{-1}}8 ms−1
  4. D
    40 ms−1\mathrm{40~ms^{-1}}40 ms−1
View written solutionFree

Correct answer: D

  1. The position (distance travelled) is given as

x=4t2x = 4t^2x=4t2

  1. Velocity is the time derivative of position:

v=dxdtv = \frac{dx}{dt}v=dtdx​

  1. Differentiate x=4t2x = 4t^2x=4t2 with respect to ttt:

v=ddt(4t2)=8tv = \frac{d}{dt}(4t^2) = 8tv=dtd​(4t2)=8t

  1. At t=5 st = 5\,\text{s}t=5s,

v=8(5)=40 m s−1v = 8(5) = 40\,\text{m s}^{-1}v=8(5)=40m s−1

  1. Therefore, the correct option is:

D: 40 m s−1\boxed{\text{D: }40\,\text{m s}^{-1}}D: 40m s−1​

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