Chapter 3 | Derivatives
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v ( t ) = s ′( t ) =3 t 2 - 4 and a ( t ) = v ′( t ) = s ″( t ) =6 t . Evaluating these functions at t =1, we obtain v (1) =−1 and a (1) =6. a. Because v (1) <0, the particle is moving from right to left.
b. Because v (1) <0 and a (1) >0, velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing | v ( t ) | to decrease. The particle is slowing down.
Example 3.36 Position and Velocity
The position of a particle moving along a coordinate axis is given by s ( t ) = t 3 −9 t 2 +24 t +4, t ≥0. a. Find v ( t ). b. At what time(s) is the particle at rest? c. On what time intervals is the particle moving from left to right? From right to left? d. Use the information obtained to sketch the path of the particle along a coordinate axis.
Solution a. The velocity is the derivative of the position function:
v ( t ) = s ′( t ) =3 t 2 −18 t +24. b. The particle is at rest when v ( t ) =0, so set 3 t 2 −18 t +24=0. Factoring the left-hand side of the equation produces 3( t −2)( t −4) =0. Solving, we find that the particle is at rest at t =2 and t =4. c. The particle is moving from left to right when v ( t ) >0 and from right to left when v ( t ) <0. Figure 3.23 gives the analysis of the sign of v ( t ) for t ≥0, but it does not represent the axis along which the particle is moving.
Figure 3.23 The sign of v(t) determines the direction of the particle.
Since 3 t 2 −18 t +24>0 on [0, 2) ∪ (2, +∞), the particle is moving from left to right on these intervals. Since 3 t 2 −18 t +24<0 on (2, 4), the particle is moving from right to left on this interval. d. Before we can sketch the graph of the particle, we need to know its position at the time it starts moving ⎛ ⎝ t =0) and at the times that it changes direction ( t =2, 4). Wehave s (0) =4, s (2) =24, and s (4) =20. This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and
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