The position of a point in time ‘t’ is given by x = a + bt–ct 2 , y = at + bt 2 . Its…
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The position of a point in time ‘t’ is given by x = a + bt–ct2, y = at + bt2. Its acceleration at time ‘t’ is
- b – c
- b + c
- 2b – 2c
- \(2 \sqrt{b^{2}+c^{2}}\)
Answer
(D) 2 b^ 2 +c^ 2
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