The maximum velocity in the positive □-direction is (that the velocity is increasing) and negative after □ (that the velocity is decreasing), which corresponds Or by checking that the sign of the acceleration (the first derivative of the velocity) is positive before □ We need to check now that □ ( 2 ) is a maximum by either checking values of the velocity before The derivative of the velocity (i.e., the acceleration) and finding at what time it is zero, which corresponds to an extremum of The second method is more general as it works with any type of functions, not only with quadratics. The coordinates of the parabola’s vertex are ( 2, 1 4 ) this means that the velocity is at In the so-called vertex form □ = □ ( □ − ℎ ) + □ , where ( ℎ, □ ) are theĬoordinates of the parabola’s vertex. The first one consists in rewriting its equation There are two methods to find the coordinates of a parabola’s vertex. The vertex of the parabola corresponds to a maximum value of the velocity. Its graph is therefore a parabola opening downward. The velocity function is a quadratic function with a negative leading coefficient ( − 3 ). Find the maximum velocity of the particle in the positive Our findings can be easily checked by graphing □ ( □ ), which is a parabola, as seen in theĮxample 2: Using Derivatives to Find the Maximum Velocity of a Particle given Its Displacement FunctionĪ particle is moving in a straight line along the □-axis such that its displacement In ourĬase, the velocity is negative and thus decreases further. The body is then accelerated (here in the negative direction) and the speed increases. When the acceleration and velocity are of opposite signs, the body isĪctually decelerating, which means that its speed decreases until it reaches zero, at which point the body changes direction It is consistent with an acceleration being negative. In the previous example, it is worth noting that in the time interval found, the velocity goes from positive to negative Of the particle is decreasing for values of □ in this interval. The acceleration is therefore negative for 0 ≤ □ < 5 5 1 2 s e c o n d s, which means that the velocity sįor □ < □ , □ ( □ ) < 0 this can be seen from the fact that theĪcceleration is a linear function with a positive slope (and is therefore increasing), or simply by looking at the sign We see that the acceleration function is a linear function that is equal to zero at □ such thatĢ 4 □ − 1 1 0 = 0 2 4 □ = 1 1 0 □ = 1 1 0 2 4 □ = 5 5 1 2. The acceleration is the derivative of the velocity with respect to time: To find the time interval for which □ ( □ ) is decreasing, we need to study the sign of theĭerivative of □ ( □ ) that is, of the acceleration function □ ( □ ). d d d d m s Īt the instant □ = 8, the value of □ is given by The derivative of □ ( □ ) with respect to □ is □ ( □ ), The displacement of the particle can be expressed as a function of time, □ ( □ ). Find the time interval during which the velocity of the particle is decreasing.Find the velocity of the particle at □ = 8 s e c o n d s.Example 1: Using Derivatives to Solve Problems Involving Displacement and VelocityĪ particle is moving in a straight line such that its displacement
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