📖 generic · 12th TN - English Medium · MATHEMATICS-VOLUME 2 · Page 7definition

7.2.2 Derivative as rate of change · Part 4

Chapter 3: Chapter 7 · MATHEMATICS-VOLUME 2

the velocity v ( t ) of the particle is zero. Now, we find the velocity of the particle at time t . v t ds dt ( ) = v t ( ) = ⇒ ⇒= . At t = seconds, the particle reaches the maximum height.

The height at t = is s ( ) ft. (ii) When the particle hits the ground then s = . s ⇒ ⇒ seconds. The particle hits the ground at t = seconds.

The velocity when it hits the ground is v ( ) = – ft /s. Example . A particle moves along a horizontal line such that its position at any time t ≥ is given by s t ( ) = , where s is measured in metres and t in seconds? (i) At what time the particle is at rest?

(ii) At what time the particle changes its direction? (iii) Find the total distance travelled by the particle in the first seconds. Given that s t ( ) = . On differentiating, we get v t ( ) = and a t (i) The particle is at rest when v t ( ) = .

Therefore, v t )( gives t = and t = . (ii) The particle changes its direction when v t ( ) changes its sign. Now. if ≤< then both ( t − and ( t − and hence, v t ( ) > .

If < < then ( t − > and ( t − and hence, v t ( ) < . If t > then both ( t − and ( t − > and hence, v t ( ) > . Therefore, the particle changes its direction when t =

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