MHT CET · Physics · Motion In One Dimension
Two bodies ' \(A\) ' and ' \(B\) ' start from the same point at the same instant and move along a straight line. 'A' moves with uniform acceleration ‘A’ and 'B' moves with uniform velocity (V). They meet after time ' \(t\) '. value of ' \(t\) ' is
- A \(\frac{2 \mathrm{~V}}{\mathrm{a}}\)
- B \(\sqrt{\frac{V}{a}}\)
- C \(\frac{\mathrm{a}}{2 \mathrm{~V}}\)
- D \(\frac{\mathrm{V}}{2 \mathrm{a}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 \mathrm{~V}}{\mathrm{a}}\)
Step-by-step Solution
Detailed explanation
For body \(\mathrm{B}\), distance travelled is \(\mathrm{x}=\mathrm{Vt}\)
For body \(\mathrm{A}, \mathrm{x}=\frac{1}{2} \mathrm{at}^2\)
\(\begin{aligned}& \Rightarrow \frac{1}{2} \mathrm{at}^2=\mathrm{Vt} \\& \text { Or } \mathrm{t}=\frac{2 \mathrm{~V}}{\mathrm{a}}\end{aligned}\)
For body \(\mathrm{A}, \mathrm{x}=\frac{1}{2} \mathrm{at}^2\)
\(\begin{aligned}& \Rightarrow \frac{1}{2} \mathrm{at}^2=\mathrm{Vt} \\& \text { Or } \mathrm{t}=\frac{2 \mathrm{~V}}{\mathrm{a}}\end{aligned}\)
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