MHT CET · Physics · Motion In One Dimension
Two bodies 'A' an 'B' start from the same point at the same instant and move along a straight line. Body 'A' moves with uniform acceleration ' \(a\) ' and body ' \(B\) ' moves with uniform velocity ' \(V\) '. They meet after time ' \(t\) '. The value of ' \(t\) ' is
- A \(\frac{2 \mathrm{~V}}{\mathrm{a}}\)
- B \(\frac{a}{2 V}\)
- C \(\frac{\mathrm{V}}{2 \mathrm{a}}\)
- D \(\sqrt{\frac{V}{a}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{V}}{2 \mathrm{a}}\)
Step-by-step Solution
Detailed explanation
At point \(A\),
\(\mathrm{S}_{\mathrm{A}}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\)
\(\mathrm{S}_{\mathrm{A}}=\frac{1}{2} \mathrm{at}^2\)
At point \(B\),
\(\mathrm{S}_{\mathrm{B}}=\mathrm{vt}\)
\(\therefore \quad\) At the point where they meet,
\(\mathrm{S}_{\mathrm{A}}=\mathrm{S}_{\mathrm{B}}\)
\(\frac{1}{2} a t^2=v t\)
\(t=\frac{2 v}{a}\)
\(\mathrm{S}_{\mathrm{A}}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\)
\(\mathrm{S}_{\mathrm{A}}=\frac{1}{2} \mathrm{at}^2\)
At point \(B\),
\(\mathrm{S}_{\mathrm{B}}=\mathrm{vt}\)
\(\therefore \quad\) At the point where they meet,
\(\mathrm{S}_{\mathrm{A}}=\mathrm{S}_{\mathrm{B}}\)
\(\frac{1}{2} a t^2=v t\)
\(t=\frac{2 v}{a}\)
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