
Solve: The posistion of a particle moving along a coordinate line is s=sqrt(5+4t), with s in meters and t in seconds. Find the particle's velocity at t=1 sec. A) 2/3 m/sec B) 4/3 m/sec C) 1/3 m/sec D) 1/6 m/sec Thank you!

Me again. One last question! Again, just needed my answer verified with any explanation or walkthrough. The position of a particle moving along a coordinate line is s=√(3+6t) with s in meters and t in seconds. find the particle's acceleration at t=1

Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? thanks. = ¡ì sec x d tan x = sec x tan x  ¡ì tan x d sec x = sec x tan x  ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx  ¡ì sec^3(x) dx = sec x tan x

a particle moves along a number line measured in cm so that its position at time t sec is given by s=72/(t+2) +k, k is a constant and t>=0 seconds. (a) Find the instantaneous velocity of the particle at t=4 seconds (b) Find the acceleration of the

A particle moves along a line so that its position at any time t >= 0 is given by the function t^3 + t^2 + 5t + 3, where p is measured in feet and t is measured in seconds. 1. Find the displacement during the first four seconds. My answer: 75 ft 2.


The magnitude of the velocity of a particle which starts from rest 2 ft below the origin when t = 0 and moves along a vertical axis is directly proportional to the time after starting. The displacement of the particle during the time interval from t = 1

The magnitude of the velocity of a particle which starts from rest 2 ft below the origin when t = 0 and moves along a vertical axis is directly proportional to the time after starting. The displacement of the particle during the time interval from t = 1

2 given the curve is described by the equation r=3cos ¥è, find the angle that the tangent line makes with the radius vector when ¥è=120¨¬. A. 30¨¬ B. 45¨¬ C. 60¨¬ D. 90¨¬ not sure A or D 2.) which of the following represents dy/dx when

1.) which of the following represents dy/dx when y=e^2x Sec(3x)? A.3e^2x sec(3x) tan (3x)2e^2x Sec(3x)<<<< my choice B.)3e^2x sec(3x)tan (3x)2xe^2x sec(3x) C.)3e^2x sec(3x)tan (x)2e^2x Sec(3x) D.)3e^2x sec(3x)tan (x)2xe^2x sec(3x)

the position of a particle moving along a coordinate line is s=√(1+4t) , with s in metres and t in seconds. Find the particle's velocity and acceleration at t=6 sec

1.) which of the following represents dy/dx when y=e^2x Sec(3x)? A.3e^2x sec(3x) tan (3x)2e^2x Sec(3x)<<<< my choice B.)3e^2x sec(3x)tan (3x)2xe^2x sec(3x) C.)3e^2x sec(3x)tan (x)2e^2x Sec(3x) D.)3e^2x sec(3x)tan (x)2xe^2x sec(3x)

The following table gives the velocity v(t) (in feet per sec) at different instances of time t (in sec) of a particle moving along a horizontal axis. t 0 4 8 12 16 20 v(t) 43 42 40 35 25 5 Estimate the distance traveled by the particle between t = 0 sec

A particle is put inside an accelerator at time t=0. After t sec, its velocity is 10^5t^2 m/s. How far does the particle move during the first 10**2 sec? distance= int velocity dt = INT 1E5 t^2 dt limits 0 to 1E2 sec

From the below information Speed(m/sec) 2m at 0 sec, 4m at 2 sec, 6m at 3 sec, 8m at 4 sec, 10m at 5 sec. Calculate acceleration. Distance (I am getting answer as 27m but as per book it is 30m. Here as per table velocity is not uniform.) Plz confirm me the

Find secx if sinx = 4/5 and 270 < x < 360. tan^2x+1=sec^2x (4/5)^2+1=sec^2x 16/25+1=sec^2x 17/25=sec^2x sqrt 17/5 I don't know if it should be positive or negative.


If a ball is launched into the air vertically with a velocity of 20 m/sec, how fast will it be moving after two seconds? (The gravity's acceleration is rounded to 10 m/sec/sec.)

t(sec) 1 1.5 2 2.5 v(ft/sec) 12.2 1.3 13.4 13.7 velocity of an object moving along a line at various times. How do I estimate the object's acceleration(in ft/sec^2) at t=1 TIA t(sec) 1, 1.5, 2, 2.5 v(ft/sec) 12.2 ,1.3 ,13.4 ,13.7 The second velocity of

A sprinter completes the 100 yard dash in 9.30 seconds. How long would it take her to complete 100. meters if she continued at the same speed? 1. 10.2 sec 2. 10.9 sec 3. 10.7 sec 4. 9.30 sec

Here's my question. Please walk me through it so I can fully understand it. Jon begins jogging at a steady 3 meters/sec down the middle of Lane #1 of a public track. Laaura starts even with him in the center of Lane #2 but moves at 4 meters/sec. At the

Here's my question. Please walk me through it so I can fully understand it. Jon begins jogging at a steady 3 meters/sec down the middle of Lane #1 of a public track. Laaura starts even with him in the center of Lane #2 but moves at 4 meters/sec. At the

Plz help me plot on graph and also the solution. Q. When velocity is 20 m/s time is 0 and velocity 20 m/s time is 5 sec, velocity 10 m/sec time 7 sec, velocity 20 m/sec time 10 sec and velocity 0 time 15 sec. Calculate total distance showing each step and

A proton (mass=1.67*10^27 kg) moves with a velocity of 6.00*10^6 m/sec. Upon colliding with a stationary particle of unknown mass, the proton rebounds upon its own path with a velocity of 4.00*10^6 m/sec. The collision sends the unknown particle forward

A proton (mass=1.67*10^27 kg) moves with a velocity of 6.00*10^6 m/sec. Upon colliding with a stationary particle of unknown mass, the proton rebounds upon its own path with a velocity of 4.00*10^6 m/sec. The collision sends the unknown particle forward

Use the following table to determine the average rate during the period 800 to 1200 seconds for the decomposition of H2O2. 2 H2O2(l) → 2 H2O(l) + O2(g) Time (sec) (H2O2)M 0 2.32 400 1.72 800 1.30 1200 0.98 1600 0.73 a] 1.2 x 103M/sec b] 1.6 x 103

find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x dy/dx = dy/du*du/dx


Predict when a fast toy car can pass a slower toy car . I marked off a distant of 1.5 meters and recorded the time it took the cars to reach the finish line. How do I create a graph and make a prediction? The fast car times were .87 sec. ,.94 sec. ,and .97

a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate a)acceleration in first 10 sec.

a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate a)acceleration in first 10 sec.

a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate a)acceleration in first 10 sec.

Starting from 130 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is traveling as a constant rate of 25 feet per second. The distance between the bicycle and the checkpoint is given by the equation d=13025t. At

Starting from 130 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is traveling as a constant rate of 25 feet per second. The distance between the bicycle and the checkpoint is given by the equation d=13025t. At

starting from 105 feet away a person on a bicycle rides towards a checkpoint and then passes it. the rider is traveling at a constant rate of 35 feet per second. The distance between the bicycle and the checkpoint is given by the equation d= [10535t] at

Start from 110 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is travelling at a constant rate of 30 feet per second. The distance between the bicycle and the checkpoint is given by the equation d=11030t at

the height in feet of a rocket from ground level is given by the function f(t)= 16t^2+160t. what is the instantaneous velocity of the rocket 3 seconds after it is launched? A 32 feet/sec B 64 feet/sec C 12 feet/sec D 10 feet/sec E 8 feet/sec

Starting from 105 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is traveling at a constant rate of 35 feet per second. The distance between the bicycle and the checkpoint is given by the equation d = 105  35t.


Acar is moving with acceleration 200 m in 50 sec reached to velocity 10 m\sec .what the total distance the car moved in 60 sec.

Acar is moving with acceleration 200 m in 50 sec reached to velocity 10 m\sec .what the total distance the car moved in 60 sec.

1\ Acar is moving with acceleration 200 m in 50 sec reached to velocity 10 m\sec .what the total distance the car moved in 60 sec

1\ Acar is moving with acceleration 200 m in 50 sec reached to velocity 10 m\sec .what the total distance the car moved in 60 sec

a ball is thrown upward at a speed of 12 m/s. It will reach the top of its path in about: a) 0.6 sec b) 1.2 sec c) 1.8 sec d) 2.4 sec Please explain how you find the answer!!! I cannot figure it out

I'm having some trouble with my algebra homework. Please walk me through the question completely and give me the answer so I can fully understand it if I would ever need to solve a similar problem. Here's the problem: #1: Jon begins jogging at a steady 3

Integrate: dx/sqrt(x^29) Answer: ln(x + sqrt(x^2  9)) + C I'm getting the wrong answer. Where am I going wrong: Substitute: x = 3 * sec t sqrt(x^2  9) = sqrt(3) * tan t dx = sqrt(3) * sec t * tan t Integral simplifies to: sec t dt Integrates to: lnsec

I'm having some trouble with my algebra homework. Please walk me through the question completely and give me the answer so I can fully understand it if I would ever need to solve a similar problem. Here's the problem: #1: Jon begins jogging at a steady 3

an object is moving up and inclined plane its velocity changes from 15 cm\sec to 10 cm\sec in 2 seconds .what is its acceleration solve

The time (sec) that it takes a librarian to locate an entry in a file of records on checkedout books has an exponential distribution with lambda symbol=0.5. a)What proportion of all location times are less than 20 sec? At most 20 sec? Atleast 25 sec/


A point moving on a straight horizontal line with an initial velocity of 60 fps to the right is given an acceleration of 12 fps2 to the left for 8 sec. Determine (a) the total distance traveled during the 8sec interval; (b) the displacement during the

The position function of an object moving along a straight line is given by s = f(t). The average velocity of the object over the time interval [a, b] is the average rate of change of f over [a, b]; its (instantaneous) velocity at t = a is the rate of

could anybody please explain how sec x tan x  ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx  ¡ì sec^3(x) dx What I don't understand about your question is what is ¡ì ? i just want to know if those two equations are equal, if yes, how did one

How do I find the critical values? y= 4/x + tan(πx/8) What I did is I simplified it to y= 4x^1 + tan(πx/8) then I took the derivative y'= 4x^2 + (π/8)(sec(πx/8))^2 Then I simplied it y'= 4/x^2 + (π/8)(sec(πx/8))^2 then I

An object with an inital velocity of 20 cm/sec is decelerated at 2 cm/sec/sec for 5 seconds,What will the resultant velocity be?

Sorry this is really long. Just wondering how I would do each of these A particle is moving with velocity v(t) = t^2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds

The velocity of a particle moving along the xaxis is given by f(t)=62t cm/sec. Use a graph of f(t)to find the exact change in position of the particle from time t=0 to t=4 seconds.

A car starting from rest and moving with variable acceleration possess average velocities 5m/sec, 10m/sec, 15m/sec in the first seconf and third seconds. What is the total distance covered by the car in these three seconds?

I'm doing trigonometric integrals i wanted to know im doing step is my answer right? ∫ tan^3 (2x) sec^5(2x) dx =∫ tan^2(2x) sec^4(2x) tan*sec(2x) dx =∫ (sec^2(2x)1)sec^4 tan*sec(2x) dx let u=sec x, du= 1/2 tan*sec(2x) dx =1/2∫

How do I derive the secant reduction rule? Integral (sec x)^n dx = Integral (sec x)^(n2) * (sec x)^2 dx = Integral ((tan x)^2 + 1)^(n/21) * (sec x)^2 dx Doing a substitution with: u = tax x du = (sec x)^2 dx = Integral (u^2 + 1)^(n/21) * du At this


∫ dx/ (x^2+9)^2 dx set x = 3tan u dx = 3 sec^2 u du I = 3 sec^2 u du / ( 9 tan^2 u + 9)^2 = 3 sec^2 u du / ( 81 ( tan^2 u + 1)^2 = sec^2 u du / ( 27 ( sec^2 u )^2 = du / ( 27 sec^2 u = 2 cos^2 u du / 54 = ( 1 + cos 2u) du / 54 = ( u + sin 2u / 2) /

The position of a function of a moving particle is s(t)=5+4tt^2 for 0<t<10 where s is in meters and t is measured in seconds. What is the maximum speed in m/sec of the particle on the interval [0,10]?

(a)find antiderivatives for the following functions : (i) e^5x sinh3x + 4x+6/x^2+3x+5 . (ii)root3x(x^22/x+1). (b)Evaluate the following integrals over the given intervals: (i)4/2x1  3/x+4 over [1,3] (ii)cosh 3xsinh4x over [0, ln2] (c)A particle is

weloo (a)find antiderivatives for the following functions : (i) e^5x sinh3x + 4x+6/x^2+3x+5 . (ii)root3x(x^22/x+1). (b)Evaluate the following integrals over the given intervals: (i)4/2x1  3/x+4 over [1,3] (ii)cosh 3xsinh4x over [0, ln2] (c)A particle

heeeeelp meeeee ... (a)find antiderivatives for the following functions : (i) e^5x sinh3x + 4x+6/x^2+3x+5 . (ii) ã3x(x^22/x+1). (b)Evaluate the following integrals over the given intervals: (i)4/2x1  3/x+4 over [1,3] (ii)cosh 3xsinh4x over [0, ln2]

convert the following times from seconds to minutes... 510 sec. 270 sec. 450 sec.

Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)1) dx ∫ tan(x)sec(x)[sec^4(x)sec^2(x)] dx ∫

I need help constructing a graph showing and predicting when two toy cars traveling at different speeds when the fast car can pass the slower car.The fast car traveled 1.5 meters in .87 sec.and .94 sec and .97 sec in three trials. The slow car traveled 1.5

When a ball is thrown vertically upward into the air with a velocity of 79 ft/sec its height, y(t), in feet after t seconds is given by y(t) = 79t − 16t 2 . Find the average velocity of the ball over the interval from 3 to 3 + h seconds, h 6=/= 0. 1.

A line rotates in a horizontal plane according to the equation theta=2t^3 6t,where theta is the angular position of the rotating line, in radians ,and t is the time,in seconds. Determine the angular acceleration when t=2sec. A.) 6 radians per sec^2 B.) 12


Determine all the possible values of x where 0 deg is more than or equal to x and x is more or equal to 360 deg such that i'm not sure the solution, please correct it and how to find the x value, 2 tan x  1 = cot x solution: (sec x 1)(sec x + 2) = 0 sec

1. A particle moves along the xaxis, it's position at timer given by x(t)=t/(1+t^2), t greater than or equal to 0,where t is measured in seconds and x in meters. a) find the velocity at time t. I am a little confused.. Do I find the derivative by using

I don't understand what the graph is suppose to show and what the integral equation is suppose to mean. How would I use the graph to answer the questions? f is the differentiable function whose graph is shown in the figure. The position at time t (sec) of

A particle starts moving from rest with uniform acceleration.It travels a distance x in first 2 sec and a distance y in the next 2 sec.Then Answer) y=3x

12. A particle starts moving from rest with uniform acceleration. it travels a distance x in first 2 sec and a distance y in the next 2 sec .Then options: 1.y=x 2.y=3x

i need to integrate: (secx)^4 dx let u = sec x dv =sec^3 x dx Start with this. Then, you will have to deal with the integral of sec. You should be able to solve it after a few steps. Looks a little messy.

Did I do this problem right? Find the first and second derativesimplify your answer. y=xtanx y'= (x)(sec^2 x)+(tanx)(1) y'= xsec^2 x + tanx y"= (x)(2secx)(secxtanx)+sec^2 x + sec^2 x y"=2xsec^2 x tanx + 2 sec^2 x

What is a simplified form of the expression (sec^3 theta)  (sec theta/cot^2 theta)? a. 0 b. sec thetatan theta c. cos theta d. sec theta***** Every time I attempt to work out this problem I end up with one. I know the answer is sec theta, but I do not

car accelerates from 0 to 60 mi/hr in a time of 5.8 secs. a. What is the acceleration of car in m/s squared? need help setting up the problem, thanks acceleration = (speed change)/(time interval) To get the answer in m/s^2, the speed change of 60 mi/hr

the motion of a particle along a straight line is described by the function x=(2t3)^2 where x is in metres and t is in seconds. A)find the position ,veocity and acceleration at t=2 sec. B) find the velocity of the particle at origin.


the change in velocity takes place in 10 sec interval. What is the magnitude, the algebraic sign and the direction of the average acceleration in each interval? At the beginning of the interval a body is moving to the right along the xaxis at 5 ft/sec,

A ball of 250g hits the floor at a velocity of 2,50 m/s at an angle of 70* relative to the vertical. The vertical force in function with time between the floor and the ball is: from 0 to 50 N : from 0 to 1 sec. from 50 N to 100 N : from 1 to 2 sec.

A ball of 250g hits the floor at a velocity of 2,50 m/s at an angle of 70* relative to the vertical. The vertical force in function with time between the floor and the ball is: from 0 to 50 N : from 0 to 1 sec. from 50 N to 100 N : from 1 to 2 sec.

If an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h=16t^2+vt+s(where h and s are in ft, t is in seconds and v is in ft/sec) A package is thrown downward with an

Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3  integral of(1/3)tan(3x)dx  (1/3)[ln(sec(3x))/3]  [ln(sec(3x))/9]

Jon begins jogging at a steady 3m/sec down the middle of lane one of a public track. Laura starts even with him in the center of lane two but moves 4m/sec. At the instant they begin, Ellie is located 100 meters down he track in lane four, and is heading

I am having trouble with this problem. sec^2(pi/2x)1= cot ^2x I got : By cofunction identity sec(90 degrees  x) = csc x secx csc1 = cot^2x Then split sec x and csc1 into two fractions and multiplied both numerator and denominators by csc and got: sec

Need equation: Mass of 4.24 kg acted on by external force that reduces velocity from 8.5 m/sec to 4.1 m/sec in 3.0 sec. Find magnitude of the retarding force?

a body moves with a uniform velocity of 2m/sec for 5 sec then velocity was uniformly accelerated and a velocity of 10m/sec was achieved in next 5 sec.then brakes were applied uniformly and body comes to rest in next 10 sec.calculate a)acceleration and

you walk up the aisle in a bus moving 10m/sec and a seated friend you are walking 1.5m/sec. another friend who just got off the bus standing there waving at you says you are moving 11.5m/sec.and the stalker friend riding his bike next to the bus say your


For a ballistics study a 1.9 gram bullet is fired into soft wood at 380 meters/sec and penetrates 15 cm. a.) find the retarding force (resistance) in N. b.) the time needed to stop the bullet in sec. c. acceleration of the bullet in meters/sec.

Differentiate to find critical numbers and leave function in fully factored form. g(x) = (x^2+1)^5(x^2+2)^6 g'(x) = (x^2+1)^5[6(x^2+2)^5(2x)] + (x^2+2)^6[5(x^2+1)^4(2x)] g'(x) = 2x(x^2+1)^4(x^2+2)^5[6(x^2+2)(2x) + 5(x^2+1)(2x)] g'(x) =

starting from 105 feet away a person on a bicycle rides towards a checkpoint and then passes it. the rider is traveling at a constant rate of 35 feet per second. THe distance between the bicycle and the checkpoint is given by the equation d= [10535t] at

Starting from 105 feet away a person on a bicycle rides towards a checkpoint and then passes it. the rider is traveling at a constant rate of 35 feet per second. THe distance between the bicycle and the checkpoint is given by the equation d= [10535t] at

Consider the hyperbola y = 1/x and think of it as a slide. A particle slides along the hyperbola so that its xcoordinate is increasing at a rate of f(x) units/sec. If its ycoordinate is decreasing at a constant rate of 1 unit/sec, what is f(x)? I

a car traveled at uniform velocity of 20m/sec for 5 sec the brakes are applied and car retards uniformly and come to rest in further 8 sec.find a)distance traveled in first five sec b)distance traveled after breaks are applied c)total displacement during

A point moves along the xaxis, and its xcoordinate after t sec is x = 4t + 16. (Assume that x is in centimeters.) (a) What is the velocity? in cm/sec

Hi! Thank you very much for your help I'm not sure what the answer to this is; how do I solve? Find antiderivative of (1/(x^2))[sec(1/x)][tan(1/x)]dx I did integration by parts and got to (1/(x^2))[sec(1/x)] + 2*[antiderivative of (1/(x^3))(sec(1/x))dx]

Posted by weloo_volley on Wednesday, April 25, 2012 at 12:16am.heeeeelp meeeee ... (a)find antiderivatives for the following functions : (i) e^5x sinh3x + 4x+6/x^2+3x+5 . (ii) �ã3x(x^22/x+1). (b)Evaluate the following integrals over the

i'm still getting this question wrong. please check for my errors: Use Simpson's Rule with n = 10 to estimate the arc length of the curve. y = tan x, 0 <or= x <or= pi/4 .. this is what i did: y' = sec(x)^2 (y')^2 = [sec(x)^2]^2 [f'(x)]^2 = sec(x)^4


I am trying to find: dy/dx for y = sec(tan x) I have the answer, but I have no idea how to get there. I know that the derivative of sec x = sec x tan x and that the derivative of tan x is sec^2 x. But sec doesn't have an x, so ...?

A record player takes 4.5 sec to spin up to 45 RPM. What is the angular displacement and angular velocity when t=3 sec. I was only able to find angular acceleration from the information given. Which is 1.05 rad/sec^2

A wheel rotating with an angular velocity of 16rad/sec is slowed to an angular velocity of 4rad/sec by an angular acceleration of 2rad/sec^2.find a. The angle turned in the process of slowing down b. The time required

The velocity of a particle moving along the taxis is given by f(t)=2+.05t cm/sec. Use a graph of y=f(t) to find the exact change in position (distance traveled) for the particle from t=2 to t=8.

A bomb is dropped from a plane flying at constant horizontal velocity at an altitude of 600m. Where is the bomb after 2 sec? 5 sec? 10 sec? what are the velocities said period?