How To Find Final Velocity With Distance And Acceleration

Learning Objectives

By the end of this section, you volition be able to:

Derive the kinematic equations for constant acceleration using integral calculus.
Use the integral formulation of the kinematic equations in analyzing motion.
Observe the functional grade of velocity versus fourth dimension given the acceleration function.
Detect the functional form of position versus time given the velocity function.

This section assumes you take enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Boilerplate and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position office we institute the velocity function, and besides by taking the derivative of the velocity office we plant the dispatch part. Using integral calculus, we can piece of work backward and calculate the velocity function from the acceleration office, and the position function from the velocity role.

Kinematic Equations from Integral Calculus

Let'southward begin with a particle with an acceleration a(t) is a known role of fourth dimension. Since the fourth dimension derivative of the velocity function is acceleration,

[latex] \frac{d}{dt}five(t)=a(t), [/latex]

we can accept the indefinite integral of both sides, finding

[latex] \int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{i}, [/latex]

where C ₁ is a constant of integration. Since [latex] \int \frac{d}{dt}v(t)dt=five(t) [/latex], the velocity is given by

[latex] v(t)=\int a(t)dt+{C}_{1}. [/latex]

Similarly, the time derivative of the position part is the velocity function,

[latex] \frac{d}{dt}10(t)=five(t). [/latex]

Thus, we tin use the same mathematical manipulations we merely used and find

[latex] ten(t)=\int five(t)dt+{C}_{2}, [/latex]

where C ₂ is a 2d constant of integration.

We can derive the kinematic equations for a constant dispatch using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find

[latex] 5(t)=\int adt+{C}_{1}=at+{C}_{1}. [/latex]

If the initial velocity is v(0) = v ₀, then

[latex] {v}_{0}=0+{C}_{1}. [/latex]

Then, C ₁ = v ₀ and

[latex] v(t)={v}_{0}+at, [/latex]

which is (Equation). Substituting this expression into (Figure) gives

[latex] x(t)=\int ({five}_{0}+at)dt+{C}_{ii}. [/latex]

Doing the integration, we find

[latex] x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}. [/latex]

If x(0) = x ₀, nosotros have

[latex] {x}_{0}=0+0+{C}_{ii}; [/latex]

so, C ₂ = x ₀. Substituting back into the equation for 10(t), we finally have

[latex] x(t)={x}_{0}+{v}_{0}t+\frac{i}{two}a{t}^{2}, [/latex]

which is (Equation).

Example

Motion of a Motorboat

A motorboat is traveling at a constant velocity of 5.0 m/southward when it starts to decelerate to arrive at the dock. Its acceleration is [latex] a(t)=-\frac{1}{four}t\,\text{m/}{\text{south}}^{2} [/latex]. (a) What is the velocity function of the motorboat? (b) At what time does the velocity accomplish cypher? (c) What is the position function of the motorboat? (d) What is the displacement of the motorboat from the fourth dimension it begins to decelerate to when the velocity is zilch? (e) Graph the velocity and position functions.

Strategy

(a) To get the velocity part we must integrate and utilise initial conditions to find the constant of integration. (b) Nosotros set the velocity part equal to zero and solve for t. (c) Similarly, we must integrate to observe the position office and use initial weather condition to discover the constant of integration. (d) Since the initial position is taken to be nix, we just take to evaluate the position function at [latex] t=0 [/latex].

Solution

Nosotros accept t = 0 to be the fourth dimension when the boat starts to decelerate.

From the functional form of the acceleration we tin solve (Figure) to go v(t):

Evidence Answer

[latex] v(t)=\int a(t)dt+{C}_{ane}=\int -\frac{1}{4}tdt+{C}_{ane}=-\frac{one}{8}{t}^{2}+{C}_{1}. [/latex] At t = 0 nosotros take v(0) = 5.0 m/due south = 0 + C1, and then C1 = 5.0 m/due south or [latex] five(t)=5.0\,\text{one thousand/}\text{s}-\frac{1}{eight}{t}^{ii} [/latex].
Show Answer

[latex] v(t)=0=5.0\,\text{yard/}\text{s}-\frac{1}{8}{t}^{2}⇒t=half dozen.3\,\text{south} [/latex]
Solve (Effigy):

Testify Answer

[latex] x(t)=\int v(t)dt+{C}_{2}=\int (5.0-\frac{1}{viii}{t}^{2})dt+{C}_{2}=5.0t-\frac{1}{24}{t}^{3}+{C}_{ii}. [/latex] At t = 0, nosotros set up x(0) = 0 = x0, since we are only interested in the displacement from when the boat starts to decelerate. We have [latex] x(0)=0={C}_{2}. [/latex] Therefore, the equation for the position is [latex] x(t)=v.0t-\frac{1}{24}{t}^{iii}. [/latex]
Show Answer

Since the initial position is taken to be zero, nosotros simply accept to evaluate x(t) when the velocity is zero. This occurs at t = six.three s. Therefore, the displacement is [latex] ten(6.3)=v.0(vi.3)-\frac{1}{24}{(vi.3)}^{iii}=21.1\,\text{m}\text{.} [/latex]

Graph A is a plot of velocity in meters per second as a function of time in seconds. Velocity is five meters per second at the beginning and decreases to zero. Graph B is a plot of position in meters as a function of time in seconds. Position is zero at the beginning, increases reaching maximum between six and seven seconds, and then starts to decrease.

Figure 3.30 (a) Velocity of the motorboat as a office of fourth dimension. The motorboat decreases its velocity to zero in half-dozen.3 s. At times greater than this, velocity becomes negative—meaning, the boat is reversing direction. (b) Position of the motorboat as a function of time. At t = 6.3 south, the velocity is zip and the gunkhole has stopped. At times greater than this, the velocity becomes negative—meaning, if the boat continues to move with the aforementioned acceleration, it reverses direction and heads dorsum toward where it originated.

Significance

The acceleration office is linear in time so the integration involves simple polynomials. In (Effigy), we see that if we extend the solution beyond the point when the velocity is zero, the velocity becomes negative and the boat reverses management. This tells the states that solutions can give united states information outside our immediate involvement and we should exist careful when interpreting them.

Cheque Your Understanding

A particle starts from rest and has an acceleration function [latex] 5-10t{\text{m/southward}}^{2} [/latex]. (a) What is the velocity function? (b) What is the position role? (c) When is the velocity zilch?

Summary

Integral calculus gives us a more than complete formulation of kinematics.
If dispatch a(t) is known, we tin can utilize integral calculus to derive expressions for velocity v(t) and position ten(t).
If acceleration is constant, the integral equations reduce to (Figure) and (Figure) for motion with abiding acceleration.

Key Equations

Displacement	[latex] \text{Δ}ten={x}_{\text{f}}-{x}_{\text{i}} [/latex]
Full displacement	[latex] \text{Δ}{x}_{\text{Full}}=\sum \text{Δ}{x}_{\text{i}} [/latex]
Boilerplate velocity	[latex] \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{{x}_{two}-{x}_{1}}{{t}_{2}-{t}_{ane}} [/latex]
Instantaneous velocity	[latex] five(t)=\frac{dx(t)}{dt} [/latex]
Average speed	[latex] \text{Average speed}=\overset{\text{–}}{due south}=\frac{\text{Full altitude}}{\text{Elapsed time}} [/latex]
Instantaneous speed	[latex] \text{Instantaneous speed}=\|five(t)\| [/latex]
Boilerplate acceleration	[latex] \overset{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}} [/latex]
Instantaneous acceleration	[latex] a(t)=\frac{dv(t)}{dt} [/latex]
Position from boilerplate velocity	[latex] x={ten}_{0}+\overset{\text{–}}{five}t [/latex]
Boilerplate velocity	[latex] \overset{\text{–}}{v}=\frac{{v}_{0}+v}{2} [/latex]
Velocity from acceleration	[latex] v={v}_{0}+at\enspace(\text{abiding}\,a\text{)} [/latex]
Position from velocity and acceleration	[latex] 10={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}\enspace(\text{abiding}\,a\text{)} [/latex]
Velocity from distance	[latex] {5}^{two}={v}_{0}^{2}+2a(10-{x}_{0})\enspace(\text{constant}\,a\text{)} [/latex]
Velocity of gratis fall	[latex] v={5}_{0}-gt\,\text{(positive upward)} [/latex]
Height of gratuitous fall	[latex] y={y}_{0}+{v}_{0}t-\frac{1}{ii}g{t}^{2} [/latex]
Velocity of free fall from meridian	[latex] {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0}) [/latex]
Velocity from acceleration	[latex] v(t)=\int a(t)dt+{C}_{one} [/latex]
Position from velocity	[latex] ten(t)=\int five(t)dt+{C}_{ii} [/latex]

Conceptual Questions

When given the acceleration part, what additional information is needed to notice the velocity function and position office?

Problems

The dispatch of a particle varies with fourth dimension co-ordinate to the equation [latex] a(t)=p{t}^{2}-q{t}^{3} [/latex]. Initially, the velocity and position are zero. (a) What is the velocity as a function of fourth dimension? (b) What is the position as a office of time?

Between t = 0 and t = t ₀, a rocket moves straight upward with an acceleration given by [latex] a(t)=A-B{t}^{1\,\text{/}two} [/latex], where A and B are constants. (a) If ten is in meters and t is in seconds, what are the units of A and B? (b) If the rocket starts from residuum, how does the velocity vary between t = 0 and t = t ₀? (c) If its initial position is zero, what is the rocket'due south position as a part of time during this same fourth dimension interval?

The velocity of a particle moving forth the ten-axis varies with time according to [latex] v(t)=A+B{t}^{-1} [/latex], where A = two m/southward, B = 0.25 thousand, and [latex] 1.0\,\text{s}\le t\le 8.0\,\text{southward} [/latex]. Determine the acceleration and position of the particle at t = 2.0 south and t = five.0 s. Assume that [latex] ten(t=one\,\text{south})=0 [/latex].

A particle at residue leaves the origin with its velocity increasing with time according to v(t) = 3.2t m/s. At 5.0 s, the particle'south velocity starts decreasing according to [16.0 – 1.5(t – 5.0)] yard/s. This decrease continues until t = eleven.0 south, after which the particle's velocity remains constant at 7.0 grand/s. (a) What is the acceleration of the particle equally a role of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 southward, and t = 12.0 s?

Additional Problems

Professional baseball player Nolan Ryan could pitch a baseball game at approximately 160.0 km/h. At that average velocity, how long did it accept a ball thrown by Ryan to accomplish dwelling house plate, which is 18.4 thou from the bullpen's mound? Compare this with the boilerplate reaction time of a human being to a visual stimulus, which is 0.25 south.

An plane leaves Chicago and makes the 3000-km trip to Los Angeles in 5.0 h. A second plane leaves Chicago one-half hour subsequently and arrives in Los Angeles at the same time. Compare the average velocities of the two planes. Ignore the curvature of Earth and the departure in altitude betwixt the two cities.

Unreasonable Results A cyclist rides 16.0 km due east, then 8.0 km westward, then 8.0 km east, so 32.0 km west, and finally 11.ii km eastward. If his average velocity is 24 km/h, how long did it take him to complete the trip? Is this a reasonable time?

An object has an acceleration of [latex] +ane.ii\,{\text{cm/s}}^{2} [/latex]. At [latex] t=iv.0\,\text{s} [/latex], its velocity is [latex] -three.four\,\text{cm/s} [/latex]. Determine the object'southward velocities at [latex] t=1.0\,\text{s} [/latex] and [latex] t=6.0\,\text{s} [/latex].

Show Solution

[latex] a=\frac{v-{v}_{0}}{t-{t}_{0}} [/latex], [latex] t=0,\,a=\frac{-three.4\,\text{cm/south}-{5}_{0}}{4\,\text{southward}}=1.two\,{\text{cm/southward}}^{two}⇒{v}_{0}=-eight.2\,\text{cm/s} [/latex][latex] 5={five}_{0}+at=-viii.2+1.2\,t [/latex]; [latex] 5=-vii.0\,\text{cm/southward}\enspacev=-1.0\,\text{cm/s} [/latex]

A particle moves along the 10-centrality according to the equation [latex] x(t)=2.0-4.0{t}^{2} [/latex] 1000. What are the velocity and acceleration at [latex] t=2.0 [/latex] s and [latex] t=5.0 [/latex] south?

A particle moving at constant acceleration has velocities of [latex] two.0\,\text{m/s} [/latex] at [latex] t=two.0 [/latex] s and [latex] -7.6\,\text{m/south} [/latex] at [latex] t=5.2 [/latex] southward. What is the dispatch of the particle?

Show Solution

[latex] a=-3\,{\text{m/due south}}^{ii} [/latex]

A train is moving upward a steep grade at constant velocity (come across following figure) when its caboose breaks loose and starts rolling freely forth the track. After five.0 s, the caboose is 30 chiliad behind the train. What is the acceleration of the caboose?

Figure shows a train moving up a hill.

An electron is moving in a directly line with a velocity of [latex] 4.0\,×\,{10}^{five} [/latex] k/southward. It enters a region 5.0 cm long where information technology undergoes an dispatch of [latex] 6.0\,×\,{10}^{12}\,{\text{m/s}}^{two} [/latex] along the same direct line. (a) What is the electron's velocity when it emerges from this region? b) How long does the electron take to cross the region?

An ambulance driver is rushing a patient to the hospital. While traveling at 72 km/h, she notices the traffic low-cal at the upcoming intersections has turned amber. To reach the intersection before the light turns crimson, she must travel 50 m in 2.0 s. (a) What minimum acceleration must the ambulance have to reach the intersection before the calorie-free turns red? (b) What is the speed of the ambulance when it reaches the intersection?

A motorcycle that is slowing down uniformly covers ii.0 successive km in 80 s and 120 southward, respectively. Summate (a) the acceleration of the motorbike and (b) its velocity at the beginning and finish of the 2-km trip.

Prove Solution

[latex] ane\,\text{km}={five}_{0}(80.0\,\text{southward})+\frac{1}{two}a{(80.0)}^{two} [/latex]; [latex] two\,\text{km}={v}_{0}(200.0)+\frac{1}{2}a{(200.0)}^{2} [/latex] solve simultaneously to get [latex] a=-\frac{0.one}{2400.0}{\text{km/s}}^{2} [/latex] and [latex] {v}_{0}=0.014167\,\text{km/s} [/latex], which is [latex] 51.0\,\text{km/h} [/latex]. Velocity at the cease of the trip is [latex] 5=21.0\,\text{km/h} [/latex].

A cyclist travels from point A to point B in ten min. During the starting time 2.0 min of her trip, she maintains a uniform dispatch of [latex] 0.090\,{\text{thou/s}}^{2} [/latex]. She and then travels at constant velocity for the next five.0 min. Next, she decelerates at a constant rate so that she comes to a rest at point B three.0 min later. (a) Sketch the velocity-versus-time graph for the trip. (b) What is the dispatch during the final 3 min? (c) How far does the cyclist travel?

2 trains are moving at 30 m/s in opposite directions on the same track. The engineers see simultaneously that they are on a collision course and employ the brakes when they are yard m apart. Assuming both trains have the same dispatch, what must this acceleration exist if the trains are to stop just curt of colliding?

Show Solution

[latex] a=-0.ix\,{\text{m/s}}^{two} [/latex]

A x.0-g-long truck moving with a constant velocity of 97.0 km/h passes a 3.0-m-long auto moving with a constant velocity of lxxx.0 km/h. How much time elapses betwixt the moment the front of the truck is even with the back of the automobile and the moment the back of the truck is fifty-fifty with the forepart of the machine?

A police motorcar waits in hiding slightly off the highway. A speeding car is spotted by the police automobile doing xl grand/s. At the instant the speeding machine passes the police car, the police car accelerates from residue at four m/southward² to catch the speeding auto. How long does it take the police car to grab the speeding car?

Show Solution

Equation for the speeding car: This motorcar has a abiding velocity, which is the average velocity, and is not accelerating, then use the equation for displacement with [latex] {x}_{0}=0 [/latex]:[latex] x={x}_{0}+\overset{\text{–}}{v}t=\overset{\text{–}}{5}t [/latex]; Equation for the police car: This car is accelerating, so apply the equation for displacement with [latex] {x}_{0}=0 [/latex] and [latex] {v}_{0}=0 [/latex], since the police car starts from residue: [latex] x={x}_{0}+{v}_{0}t+\frac{1}{ii}a{t}^{ii}=\frac{1}{ii}a{t}^{ii} [/latex]; Now we have an equation of move for each motorcar with a mutual parameter, which tin exist eliminated to find the solution. In this instance, we solve for [latex] t [/latex]. Step 1, eliminating [latex] x [/latex]: [latex] 10=\overset{\text{–}}{v}t=\frac{1}{2}a{t}^{2} [/latex]; Step 2, solving for [latex] t [/latex]: [latex] t=\frac{2\overset{\text{–}}{v}}{a} [/latex]. The speeding car has a constant velocity of 40 m/due south, which is its average velocity. The acceleration of the constabulary car is 4 m/s². Evaluating t, the time for the constabulary auto to reach the speeding car, we take [latex] t=\frac{2\overset{\text{–}}{five}}{a}=\frac{2(40)}{4}=20\,\text{s} [/latex].

Pablo is running in a half marathon at a velocity of 3 m/s. Another runner, Jacob, is l meters backside Pablo with the aforementioned velocity. Jacob begins to advance at 0.05 thou/southward^two. (a) How long does it take Jacob to take hold of Pablo? (b) What is the distance covered past Jacob? (c) What is the final velocity of Jacob?

Unreasonable results A runner approaches the finish line and is 75 grand away; her boilerplate speed at this position is 8 g/s. She decelerates at this point at 0.5 m/southward². How long does information technology accept her to cantankerous the terminate line from 75 m away? Is this reasonable?

Testify Solution

At this acceleration she comes to a full stop in [latex] t=\frac{-{five}_{0}}{a}=\frac{eight}{0.5}=16\,\text{due south} [/latex], but the distance covered is [latex] x=8\,\text{m/s(16}\,\text{s)}-\frac{one}{2}(0.v){(sixteen\,\text{southward})}^{ii}=64\,\text{m} [/latex], which is less than the altitude she is abroad from the finish line, so she never finishes the race.

An aeroplane accelerates at 5.0 m/due south² for xxx.0 s. During this time, it covers a altitude of x.0 km. What are the initial and final velocities of the aeroplane?

Compare the distance traveled of an object that undergoes a change in velocity that is twice its initial velocity with an object that changes its velocity by four times its initial velocity over the same time menstruum. The accelerations of both objects are constant.

An object is moving due east with a constant velocity and is at position [latex] {x}_{0}\,\text{at}\,\text{time}\,{t}_{0}=0 [/latex]. (a) With what acceleration must the object have for its total deportation to be cipher at a later fourth dimension t ? (b) What is the concrete interpretation of the solution in the example for [latex] t\to \infty [/latex]?

A ball is thrown straight upwards. Information technology passes a 2.00-m-loftier window seven.50 m off the ground on its path up and takes one.30 s to become past the window. What was the ball'southward initial velocity?

A money is dropped from a gasbag airship that is 300 one thousand above the basis and rising at 10.0 grand/due south upward. For the coin, find (a) the maximum meridian reached, (b) its position and velocity 4.00 s after being released, and (c) the fourth dimension before it hits the footing.

A soft lawn tennis ball is dropped onto a hard floor from a height of one.50 m and rebounds to a height of 1.10 yard. (a) Calculate its velocity only earlier information technology strikes the flooring. (b) Summate its velocity but after it leaves the flooring on its way back up. (c) Summate its acceleration during contact with the floor if that contact lasts three.50 ms [latex] (3.fifty\,×\,{10}^{-three}\,\text{s}) [/latex] (d) How much did the ball compress during its collision with the flooring, assuming the floor is admittedly rigid?

Unreasonable results. A raindrop falls from a cloud 100 m above the footing. Neglect air resistance. What is the speed of the raindrop when it hits the ground? Is this a reasonable number?

Compare the time in the air of a basketball histrion who jumps 1.0 grand vertically off the flooring with that of a player who jumps 0.three m vertically.

Suppose that a person takes 0.5 south to react and move his paw to catch an object he has dropped. (a) How far does the object fall on Earth, where [latex] g=ix.8\,{\text{k/s}}^{2}? [/latex] (b) How far does the object fall on the Moon, where the acceleration due to gravity is one/half dozen of that on World?

A gasbag balloon rises from ground level at a abiding velocity of 3.0 m/south. One minute after liftoff, a sandbag is dropped accidentally from the airship. Calculate (a) the time it takes for the sandbag to attain the footing and (b) the velocity of the sandbag when it hits the ground.

(a) A globe record was set for the men's 100-thou dash in the 2008 Olympic Games in Beijing past Usain Commodities of Jamaica. Bolt "coasted" beyond the finish line with a time of 9.69 due south. If nosotros presume that Bolt accelerated for 3.00 s to attain his maximum speed, and maintained that speed for the residuum of the race, summate his maximum speed and his acceleration. (b) During the same Olympics, Bolt likewise set the world tape in the 200-m dash with a fourth dimension of xix.30 due south. Using the same assumptions as for the 100-g nuance, what was his maximum speed for this race?

An object is dropped from a top of 75.0 1000 in a higher place ground level. (a) Make up one's mind the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the altitude traveled during the concluding second of motion earlier hitting the ground.

A steel brawl is dropped onto a hard flooring from a elevation of ane.50 chiliad and rebounds to a elevation of i.45 g. (a) Calculate its velocity just earlier information technology strikes the flooring. (b) Calculate its velocity just later it leaves the flooring on its way back up. (c) Summate its acceleration during contact with the floor if that contact lasts 0.0800 ms [latex] (eight.00\,×\,{10}^{-5}\,\text{south}) [/latex] (d) How much did the brawl compress during its collision with the floor, assuming the floor is absolutely rigid?

An object is dropped from a roof of a building of top h. During the last second of its descent, it drops a distance h/three. Summate the height of the building.

Challenge Problems

In a 100-m race, the winner is timed at eleven.2 south. The 2d-place finisher's time is 11.6 s. How far is the second-place finisher backside the winner when she crosses the end line? Presume the velocity of each runner is constant throughout the race.

The position of a particle moving along the 10-axis varies with fourth dimension according to [latex] x(t)=5.0{t}^{2}-4.0{t}^{three} [/latex] m. Find (a) the velocity and acceleration of the particle equally functions of time, (b) the velocity and acceleration at t = ii.0 s, (c) the fourth dimension at which the position is a maximum, (d) the time at which the velocity is aught, and (e) the maximum position.

A cyclist sprints at the terminate of a race to clinch a victory. She has an initial velocity of 11.5 m/s and accelerates at a rate of 0.500 m/south² for 7.00 due south. (a) What is her final velocity? (b) The cyclist continues at this velocity to the stop line. If she is 300 m from the finish line when she starts to accelerate, how much time did she save? (c) The second-place winner was 5.00 k alee when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.eight g/s until the finish line. What was the deviation in finish fourth dimension in seconds betwixt the winner and runner-up? How far back was the runner-up when the winner crossed the terminate line?

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 295.38 km/h. The one-way class was 8.00 km long. Dispatch rates are often described past the time it takes to reach 96.0 km/h from rest. If this fourth dimension was iv.00 southward and Burt accelerated at this charge per unit until he reached his maximum speed, how long did information technology take Burt to complete the class?