InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A particle moving in a straight line covers a distance of x cm in t second, where x = t^3 + 6t^2 – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?(a) 20cm/sec(b) 22cm/sec(c) 21cm/sec(d) 23cm/secThis question was posed to me in class test.Query is from Application of Derivative in division Application of Derivatives of Mathematics – Class 12 |
|
Answer» CORRECT OPTION is (c) 21cm/sec The explanation is: We have, x = t^3 + 6t^2 – 15t + 18 Let, V be the velocity of the particle at the END of t SECONDS. Then, v = dx/dt = d/dt(t^3 + 6t^2 – 15t + 18) So, v = 3t^2 + 12t – 15 Thus, velocity of the particle at the end of 2 seconds is, [dx/dt]t = 2 = 3(2)^2 + 12(2) – 15 = 21cm/sec. |
|