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https://www.cliffsnotes.com/study-guides/calculus/calculus/integration/distance-velocity-and- acceleration
he indefinite integral is commonly applied in problems involving distance, velocity, and acceleration, each of which is a function of time. In the discussion of the applications of the derivative, note that the derivative of a distance function represents instantaneous velocity and that the derivative of the velocity function represents instantaneous acceleration at a particular time. In considering the relationship between the derivative and the indefinite integral as inverse operations, note that the indefinite integral of the acceleration function represents the velocity function and that the indefinite integral of the velocity represents the distance function. In case of a free‐falling object, the acceleration due to gravity is –32 ft/sec 2. The significance of the negative is that the rate of change of the velocity with respect to time (acceleration), is negative because the velocity is decreasing as the time increases. Using the fact that the velocity is the indefinite integral of the acceleration, you find that Now, at t = 0, the initial velocity ( v (^) 0 ) is hence, because the constant of integration for the velocity in this situation is equal to the initial velocity, write Because the distance is the indefinite integral of the velocity, you find that
ow, at t = 0, the initial distance ( s (^) 0 ) is hence, because the constant of integration for the distance in this situation is equal to the initial distance, write Example 1: A ball is thrown downward from a height of 512 feet with a velocity of 64 feet per second. How long will it take for the ball to reach the ground? From the given conditions, you find that
After 6 seconds, you find that hence, the missile will be 109 m above the ground after 6 seconds.
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/3-6-finding-velocity-and- displacement-from-acceleration/ his section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function. Let’s begin with a particle with an acceleration a (t) is a known function of time. Since the time derivative of the velocity function is acceleration, ddtv(t)=a(t),ddtv(t)=a(t), we can take the indefinite integral of both sides, finding ∫ddtv(t)dt=∫a(t)dt+C 1 ,∫ddtv(t)dt=∫a(t)dt+C1, where C 1 is a constant of integration. Since ∫ddtv(t)dt=v(t)∫ddtv(t)dt=v(t), the velocity is given by v(t)=∫a(t)dt+C 1 .v(t)=∫a(t)dt+C1. Similarly, the time derivative of the position function is the velocity function, ddtx(t)=v(t).ddtx(t)=v(t). Thus, we can use the same mathematical manipulations we just used and find x(t)=∫v(t)dt+C 2 ,x(t)=∫v(t)dt+C2, where C 2 is a second constant of integration. We can derive the kinematic equations for a constant acceleration using these integrals. With a ( t ) = a a constant, and doing the integration in (Figure) , we find v(t)=∫adt+C 1 =at+C 1 .v(t)=∫adt+C1=at+C1. If the initial velocity is v (0) = v 0 , then v 0 =0+C 1 .v0=0+C1.
Motion of a Motorboat A motorboat is traveling at a constant velocity of 5.0 m/s when it starts to decelerate to arrive at the dock. Its acceleration is a(t)=− 1 4 tm/s 2 a(t)=−14tm/s2. (a) What is the velocity function of the motorboat? (b) At what time does the velocity reach zero? (c) What is the position function of the motorboat? (d) What is the displacement of the motorboat from the time it begins to decelerate to when the velocity is zero? (e) Graph the velocity and position functions. Strategy (a) To get the velocity function we must integrate and use initial conditions to find the constant of integration. (b) We set the velocity function equal to zero and solve for t. (c) Similarly, we must integrate to find the position function and use initial conditions to find the constant of integration. (d) Since the initial position is taken to be zero, we only have to evaluate the position function at t=0t=0. We take t = 0 to be the time when the boat starts to decelerate.
- From the functional form of the acceleration we can solve (Figure) to get v ( t ): v(t)=∫a(t)dt+C 1 =∫− 14 tdt+C 1 =− 18 t 2 +C 1 .v(t)=∫a(t)dt+C1=∫−14tdt+C1=−18t2+ C1. At t = 0 we have v(0) = 5.0 m/s = 0 + C1, so C1 = 5.0 m/s or v(t)=5.0m/s− 18 t 2 v(t)=5.0m/s−18t2.
- v(t)=0=5.0m/s− 18 t 2 ⇒t=6.3sv(t)=0=5.0m/s−18t2⇒t=6.3s
- Solve (Figure) : Show Answer
- Show Answer Since the initial position is taken to be zero, we only have to evaluate x(t) when the velocity is zero. This occurs at t = 6.3 s. Therefore, the displacement is x(6.3)=5.0(6.3)− 124 (6.3) 3 =21.1m.
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/3-4-motion-with-constant- acceleration/#eq3.
d. Since the initial position is taken to be zero, we only have to evaluate the position function at the time when the velocity is zero. This occurs at t = 6.3 s. Therefore, the displacement is x(6.3)=5.0(6.3s)−124(6.3s)=21.1m. https://www.intmath.com/applications-integration/1-apps-indefinite-integral.php
