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Motion in A Plane question

2025 · 24 Jan · Shift 2 · Q67
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  5. /2025 · 24 Jan · Shift 2 · Q67

Motion in A Plane question

2025 · 24 Jan · Shift 2 · Q67

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
The position vector of a moving body at any instant of time is given as r→=(5t2i^−5tj^)m\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}r=(5t2i^−5tj^​)m. The magnitude and direction of velocity at t=2st=2 st=2s is,
  1. A
    517 m/s5 \sqrt{17} \mathrm{~m} / \mathrm{s}517​ m/s, making an angle of tan⁡−14\tan ^{-1} 4tan−14 with - ve Y axis
  2. B
    515 m/s5 \sqrt{15} \mathrm{~m} / \mathrm{s}515​ m/s, making an angle of tan⁡−14\tan ^{-1} 4tan−14 with + ve XXX axis
  3. C
    517 m/s5 \sqrt{17} \mathrm{~m} / \mathrm{s}517​ m/s, making an angle of tan⁡−14\tan ^{-1} 4tan−14 with + ve XXX axis
  4. D
    515 m/s5 \sqrt{15} \mathrm{~m} / \mathrm{s}515​ m/s, making an angle of tan⁡−14\tan ^{-1} 4tan−14 with −-− ve YYY axis
View written solutionFree

Correct answer: A

  1. Given position vector
r⃗(t)=(5t2 i^−5t j^) m\vec r(t)=\left(5t^2\,\hat i-5t\,\hat j\right)\text{ m}r(t)=(5t2i^−5tj^​) m

So the coordinates are:

x=5t2,y=−5tx=5t^2, \qquad y=-5tx=5t2,y=−5t
  1. Find velocity vector

Velocity is the time derivative of position:

v⃗=dr⃗dt\vec v=\frac{d\vec r}{dt}v=dtdr​

Therefore,

v⃗=ddt(5t2i^−5tj^)=10ti^−5j^\vec v=\frac{d}{dt}(5t^2\hat i-5t\hat j) =10t\hat i-5\hat jv=dtd​(5t2i^−5tj^​)=10ti^−5j^​
  1. Evaluate at t=2 t=2\,t=2s
v⃗(2)=10(2)i^−5j^=20i^−5j^\vec v(2)=10(2)\hat i-5\hat j=20\hat i-5\hat jv(2)=10(2)i^−5j^​=20i^−5j^​
  1. Magnitude of velocity
∣v⃗∣=202+(−5)2=400+25=425=517 m/s|\vec v|=\sqrt{20^2+(-5)^2} =\sqrt{400+25} =\sqrt{425} =5\sqrt{17}\ \text{m/s}∣v∣=202+(−5)2​=400+25​=425​=517​ m/s
  1. Direction of velocity

The velocity components are:

  • along +X+X+X: 202020
  • along −Y-Y−Y: 555

So the vector lies in the fourth quadrant.

Angle with the +X+X+X axis is:

tan⁡θ=∣vy∣vx=520=14\tan\theta=\frac{|v_y|}{v_x}=\frac{5}{20}=\frac14tanθ=vx​∣vy​∣​=205​=41​

Thus,

θ=tan⁡−1(14)\theta=\tan^{-1}\left(\frac14\right)θ=tan−1(41​)

Now angle with the negative YYY axis say ϕ\phiϕ is:

tan⁡ϕ=vx∣vy∣=205=4\tan\phi=\frac{v_x}{|v_y|}=\frac{20}{5}=4tanϕ=∣vy​∣vx​​=520​=4

Hence,

ϕ=tan⁡−1(4)\phi=\tan^{-1}(4)ϕ=tan−1(4)

So the velocity makes an angle tan⁡−1(4)\tan^{-1}(4)tan−1(4) with the negative YYY axis.

  1. Match with options

The correct option is:

A: 517 m/s, angle tan⁡−14 with −Y axis\boxed{\text{A: }5\sqrt{17}\ \text{m/s, angle }\tan^{-1}4\text{ with }-Y\text{ axis}}A: 517​ m/s, angle tan−14 with −Y axis​
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