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Motion in A Straight Line question

2019 · 11 Jan · Shift 2 · Q50
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  5. /2019 · 11 Jan · Shift 2 · Q50

Motion in A Straight Line question

2019 · 11 Jan · Shift 2 · Q50

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
A particle moves from the point (2.0i^+4.0j^)\left( {2.0\widehat i + 4.0\widehat j} \right)(2.0i+4.0j​) m, at t = 0, with an initial velocity (5.0i^+4.0j^)\left( {5.0\widehat i + 4.0\widehat j} \right)(5.0i+4.0j​) ms −-− 1. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)\left( {4.0\widehat i + 4.0\widehat j} \right)(4.0i+4.0j​) ms −-− 2. What is the distance of the particle from the origin at time 2 s?
  1. A
    15 m
  2. B
    20220\sqrt 2202​ m
  3. C
    10210\sqrt 2102​ m
  4. D
    5 m
View written solutionFree

Correct answer: B

  1. Given data
  • Initial position: r⃗0=2i^+4j^ m\vec r_0 = 2\hat i + 4\hat j \text{ m}r0​=2i^+4j^​ m
  • Initial velocity: u⃗=5i^+4j^ m s−1\vec u = 5\hat i + 4\hat j \text{ m s}^{-1}u=5i^+4j^​ m s−1
  • Constant acceleration: a⃗=4i^+4j^ m s−2\vec a = 4\hat i + 4\hat j \text{ m s}^{-2}a=4i^+4j^​ m s−2
  • Time: t=2 st=2\text{ s}t=2 s
  1. Use vector equation of motion

Position at time ttt is

r⃗=r⃗0+u⃗t+12a⃗t2\vec r = \vec r_0 + \vec u t + \frac{1}{2}\vec a t^2r=r0​+ut+21​at2

Substitute t=2t=2t=2 s:

r⃗=(2i^+4j^)+(5i^+4j^)(2)+12(4i^+4j^)(2)2\vec r = (2\hat i+4\hat j) + (5\hat i+4\hat j)(2) + \frac{1}{2}(4\hat i+4\hat j)(2)^2r=(2i^+4j^​)+(5i^+4j^​)(2)+21​(4i^+4j^​)(2)2
  1. Compute each term
  • Velocity term:
(5i^+4j^)(2)=10i^+8j^(5\hat i+4\hat j)(2)=10\hat i+8\hat j(5i^+4j^​)(2)=10i^+8j^​
  • Acceleration term:
12(4i^+4j^)(4)=2(4i^+4j^)=8i^+8j^\frac{1}{2}(4\hat i+4\hat j)(4)=2(4\hat i+4\hat j)=8\hat i+8\hat j21​(4i^+4j^​)(4)=2(4i^+4j^​)=8i^+8j^​
  1. Add all vectors
r⃗=(2i^+4j^)+(10i^+8j^)+(8i^+8j^)\vec r = (2\hat i+4\hat j)+(10\hat i+8\hat j)+(8\hat i+8\hat j)r=(2i^+4j^​)+(10i^+8j^​)+(8i^+8j^​) r⃗=20i^+20j^\vec r = 20\hat i+20\hat jr=20i^+20j^​

So at t=2t=2t=2 s, the particle is at

(20,20)(20,20)(20,20)
  1. Distance from origin
∣r⃗∣=202+202|\vec r|=\sqrt{20^2+20^2}∣r∣=202+202​ =400+400=800=202 m=\sqrt{400+400}=\sqrt{800}=20\sqrt{2}\text{ m}=400+400​=800​=202​ m
  1. Check options
  • A: 151515 m ❌
  • B: 20220\sqrt 2202​ m ✅
  • C: 10210\sqrt 2102​ m ❌
  • D: 555 m ❌

Therefore, the correct answer is Option B.

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