JEE Mains · Physics · STD 11 - 3.2 motion in plane
The co-ordinates of a particle moving in \(x-y\) plane are given by : \(\mathrm{x}=2+4 \mathrm{t}, \mathrm{y}=3 \mathrm{t}+8 \mathrm{t}^2 .\) The motion of the particle is _______.
- A non-uniformly accelerated.
- B uniformly accelerated having motion along a straight line.
- C uniform motion along a straight line.
- D uniformly accelerated having motion along a parabolic path.
Answer & Solution
Correct Answer
(D) uniformly accelerated having motion along a parabolic path.
Step-by-step Solution
Detailed explanation
\(x=2+4 t\) \(\frac{d x}{d t}=v_x=4\) \(\frac{d v_x}{d t}=a_x=0\) \(y=3 t+8 t^2\) \(\frac{d y}{d t}=v_y=3+16 t\) \(\frac{d v_y}{d t}=a_y=16\) the motion will be uniformly accelerated motion. For path \(\mathrm{x}=2+4 \mathrm{t}\) \(\frac{(\mathrm{x}-2)}{4}=\mathrm{t}\) Put this…
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