KCET · Physics · Motion In Two Dimensions
A particle starts from the origin at \(t=0\) with a velocity of \(10 \hat{\mathbf{j}} \mathrm{ms}^{-1}\) and move in the \(x-y\) plane with a constant acceleration of \((8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-2}\). At an instant when the \(x\)-coordinate of the particle is \(16 \mathrm{~m}, y\)-coordinate of the particle is
- A \(16 \mathrm{~m}\)
- B \(28 \mathrm{~m}\)
- C \(36 \mathrm{~m}\)
- D \(24 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(24 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Given, \(\mathbf{u}=10 \hat{\mathbf{j}} \mathrm{ms}^{-1}, \mathbf{a}=8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \mathrm{ms}^{-2}\) \(x=16 \mathrm{~m}\)
Using equation of motion,
\(s=u t+\frac{1}{2} a t^{2} \)
\( x \hat{\mathbf{i}}+y \hat{\mathbf{j}}=10 \hat{\mathbf{j}} \times t+\frac{1}{2}(8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) t^{2}\)
\(\Rightarrow 16 \hat{\mathbf{i}}+y \hat{\mathbf{j}}=10 t \hat{\mathbf{j}}+4 t^{2} \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}\)
Comparing both sides, we get
\(4 t^{2}=16\)
\(\Rightarrow t=2 \mathrm{~s}\)
and \(y=10 t+t^{2}\)
\(=10 \times 2+(2)^{2}\)
\(=24 \mathrm{~m}\)
Using equation of motion,
\(s=u t+\frac{1}{2} a t^{2} \)
\( x \hat{\mathbf{i}}+y \hat{\mathbf{j}}=10 \hat{\mathbf{j}} \times t+\frac{1}{2}(8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) t^{2}\)
\(\Rightarrow 16 \hat{\mathbf{i}}+y \hat{\mathbf{j}}=10 t \hat{\mathbf{j}}+4 t^{2} \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}\)
Comparing both sides, we get
\(4 t^{2}=16\)
\(\Rightarrow t=2 \mathrm{~s}\)
and \(y=10 t+t^{2}\)
\(=10 \times 2+(2)^{2}\)
\(=24 \mathrm{~m}\)
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