WBJEE · Maths · Differentiation
A particle starts moving from rest from a fixed point in a fixed direction. The distance s from the fixed point at a timet is given by \(s=t^{2}+a t-b+17,\) where \(a\) and \(b\) are real numbers. If the particle comes to rest after 5s at a distance of \(s=25\) units from the fixed point, then values of \(a\) and \(b\) are. respectively
- A 10, -33
- B -10,-30
- C -8.33 .
- D -10.33
Answer & Solution
Correct Answer
(D) -10.33
Step-by-step Solution
Detailed explanation
Given, \(s=t^{2}+a t-b+17\) After 5 s. \(\left[\frac{d s}{d t}\right]_{t=5}=[2 t+a]_{t=5}=0\) \(\Rightarrow \quad a=-10\) At \(t=5 \mathrm{s}, s=25\) units \(\therefore \quad 25=5^{2}+5 a-b+17\) \[ \mid \because s=25 \text { and } t=5 \mid \]…
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