The acceleration is the derivative of the velocity.

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Presentation transcript:

The acceleration is the derivative of the velocity. Part (a) The acceleration is the derivative of the velocity. v(t) = 1 - tan-1(et) a(t) = -et 1 + e2t a(2) = - e2 1 + e4 a(2) = -.132 or -.133

Since both v(2) and a(2) are negative, the speed is increasing. Part (b) v(2) = 1 - tan-1(e2) v(2) = -.436 a(2) = -.133 from Part (a) Since both v(2) and a(2) are negative, the speed is increasing. Remember from earlier in the year that speed INCREASES when v(t) and a(t) have the SAME sign, and speed DECREASES when the signs are OPPOSITE.

Part (c) v(t) = 1 - tan-1(et) 0 = 1 - tan-1(et) tan-1(et) = 1 The particle reaches its highest point when v = 0. v(t) = 1 - tan-1(et) 0 = 1 - tan-1(et) tan-1(et) = 1 tan 1 = et et = tan 1 ln (et) = ln (tan 1)

Part (c) Radians required here!!! t (ln e) = ln (tan 1) t = ln (tan 1) t = .443 sec ln (et) = ln (tan 1)

Part (c) To justify that t=.443 is the highest point, we’ll find the velocity somewhere below and above t=.443 seconds. v(.3) = 1 - tan-1(e.3) = .067 t = .443 sec v(.5) = 1 - tan-1(e.5) = -.026 Since v(t) is positive for t<.443 and v(t) is negative for t>.443, the particle’s graph has an absolute maximum at t=.443

Part (d) v(t) = 1 - tan-1(et) Integrate both sides y(2) = -1 + [1- tan-1(et)] dt 2 y(2) = -1.360 or -1.361 The particle is moving AWAY from the origin since v(2) and y(2) are both negative.