A particle moves along the x-axis so that at a time t > equal to its position is given by x(t)= 2t^3-21t^2..
The velocity of the particle can be found using Calculus, specifically the first derivative. v(t)=ds/dt=6t^2-42t+72 Since you are looking for the particle at rest, velocity =0 0 =6t^2-42t+72 0=6(t^2-7t+12) 0=6(t-3)(t-4) so Velocity = 0 at t =3 and 4 seconds Hope this helps.
A particle moves along the x-axis so that its velocity at any time t greater than or equal to 0 is given by..
1. x(t) = integral (v(t) ) dt = t^3 - t^2 - t + C x(2) = 5 = 8 - 4 - 2 + C C = 3 x(t) = t^3 - t^2 - t + 3 2. average velocity = (x(3) - x(0))/(3 - 0) = (18 - 3)/3 = 5 v(t) = 3t^2 - 2t - 1 = 5 3t^2 - 2t - 6 = 0 t = (2 +/- sqrt(4 - 4(3)(-6)))/(2*3) t = (2 +/- sqrt(76))/6 t = (2 +/- 2 sqrt(19))/6 t = (1 +/- sqrt(19))/3 t = -1.12 or t = 1.79 in the interval [0, 3] t = 1.79 3. v(t) = 3t^2 - 2t - 1 = (3t + 1)(t - 1) v(t) = 0 when t = 1 on the interval [0, 3] Total distance traveled = - integral (3t^2 - 2t - 1) from t = 0 to t = 1 + integral (3t^2 - 2t - 1) from t = 1 to t = 3 = -(t^3 - t^2 - t) from t = 0 to t = 1 + (t^3 - t^2 - t) from t = 1 to t = 3 = -(-1 - 0) + (15 - -1) = 1 + 16 = 17
A particle moves along x-axis in such a way that its co-ordinate x varies with the time t according to the equation x=3-7t+8t^2. What is the initial velocity and acceleration of the particle?
Let us take all quantities in SI.Given displacement: X = 3–7t + 8t²dx/dt = velocity = -7+16tTo find initial velocity, put t =0Hence u = -7 m/s.Now acceleration, dv/dt = 16 m/s²
A particle moves along the x axis as x=u(t−2) +a(t−2) 2x=u(t−2) +a(t−2) 2. How is the acceleration of this particle equal to 2a?
Look if you mean x=u(t-2)+a(t-2)^2 Then question is right.see x=ut-2u+a(t^2 + 4 -4t) Open the bracket->x=ut -2u + at^2 + 4a - 4atDifferentiate with respect to time v=u - 0 + 2at + 0 - 4aAgain diff wrt to timea= 0 + 0 + 2a + 0 -0a = 2a Dont confuse yourself with u and a on the RHS as acceleration ans velocity . They are constants and treat them as constants Alternatively by using common sense We have from laws of motion x=ut+1/2ât^2 using â instead of a to avoid confusionHere we have x= u(t-2) + a (t-2)^2compare the equations we find 1/2â=a so á=2a
A particle moves along the x-axis so that at any time t≥0, its velocity is given by v(t)=sin (2t).......?
Position = Intgr(V)dt. x(t) = Intgr(sin2t)dt = -1/2 cos(2t) + C At t = Pi/2, x(t) = -1/2cos(Pi) + C = 1/2+C = 4; C = 3.5 x(t) = -1/2 cos(2t) + 3.5 x(0) = -1/2 cos(0) + 3.5 = -1/2 + 3.5 = 3.
A particle moves along the x-axis so that at any time t>0 its velocity is given by v(t)=tlnt-t?
a) Acceleration is the rate of change of velocity. So a(t) = v'(t) = t/t + ln(t) - 1 = ln(t). b) The particle moves to the right when the velocity is positive and to the left when it is negative--this is the convention that x increases to the right. v(t) = t(ln(t) - 1), t > 0 So this is positive if the parenthetical expression is positive. ln(t) - 1 > 0 ==> ln(t) > 1 ==> t > e. c) The minimum velocity can be found by using the derivative found in part (a). v'(t) = ln(t) The critical value is t = 1 where v'(t) = 0. Using the second derivative test v''(t) = 1/t ==> v''(1) = 1 > 0 There is a minimum at t = 1. The minimum velocity is v(1) = 1ln(1) - 1 = -1.
A particle moves along the X-axis so that, at any time t is greater than or equal to 0.?
Given: a(t) = 6t + 6; v(0) = -9; x(0) = -27 We (should) know: a(t) = v'(t) = x''(t) a) v(t) = 3t² + 6t + C v(0) = -9 = 3(0²) + 6(0) + C C = -9 Therefore v(t) = 3t² + 6t - 9 or 3(t² + 2t - 3) b) Assuming the right direction is positive movement, first find zeros of v(t) 3(t + 3)(t - 1) = 0 t = -3 or t = 1 v(-4) = 3(-4)² + 6(-4) - 9 = 15 v(0) = -9 v(2) = 3(2²) + 6(2) - 9 = 15 Therefore particle is moving to the right for t < -3 and t > 1. c) x(t) = t^3 + 3t² - 9t + D x(0) = -27 = (0)^3 + 3(0²) - 9(0) + D D = -27 Therefore x(t) = t^3 + 3t² - 9t - 27
A particle moves along the x-axis so that at time t it's position is given by x(t) = sin((pie)t²) for -1 x(t) = sin(π t²) a) dx/dt = v(t) x = sin(u) dx/du = cos(u) u = π t² du/dt = 2π t dx/dt = dx/du * du/dt = cos(u) * 2π t = 2π t cos(π t²) b) dv/dt = a(t) EDIT - my first answer was incorrect (perhaps I'm not such a genius) v(t) = uv u = 2π t v = cos(π t²) u ' = 2π v ' = -2π t sin(π t²) v ' (t) = u ' v + v ' u = 2π cos(π t²) - (2π t) 2π t sin(π t²) = 2π cos(π t²) - 4π² t² sin(π t²) c) Possible change of direction when v(t) = 0 2π t cos(π t²) = 0 t = 0 or cos(π t²) = 0 π t² = ±π/2 t² = ±1/2 t = ±1/√2 Check change of direction and not just points of inflection x(-1) < x(-1/√2) x(-0.5) < x(-1/√2) so local max at t = -1/√2 x(-1/2) > x(0) x(1/2) > x(0) so local min at t = 0 x(0.5) < x(1/√2) x(1) < x(1/√2) so local max at t = 1/√2 so particle changes direction when t = -1/√2, 0 and 1/√2 You could of course have used the second derivative test a(-1/√2) < 0 so concave down - local max a(0) > 0 so concave up - local min a(1/√2) < 0 so concave down - local max d) Particle is moving left when v(t) < 0 2π t cos(π t²) < 0 t cos(π t²) < 0 if t > 0 then cos(π t²) < 0 π t² > π/2 t² > 1/2 t > 1/√2 if t < 0 then cos(π t²) > 0 π t² > -π/2 t² > -1/2 t > -1/√2 Particle moves left when -1/√2 < t < 0 and 1/√2 < t < 1 yes that's right - I am a genius (and I can spell genius too) BTW it's "pi" not "pie" - a greek letter standing for a particular mathematical constant as opposed to a tasty treat sometimes available in blueberry PS - It's MS Genius :)
A particle moves along the x axis as [math]x = u(t-2) + a(t-2)^2[/math]. How is the acceleration of this particle equal to 2a?
x = ut - 2u + at^2 +4a -4at } opening the brackets.dx/dt = v = u + 2at - 4a } taking derivative wrt to t on both sidesdv/dt = acceleration = 2a } taking derivative wrt to t on both sidesThus, Acceleration = 2a m/s^2Also, we know thatx = ut + 1/2at^2 --- 1Comparing the given equation with the above equation we getAcceleration = 2a m/s^2Note: for converting equation 1 in the form of the given equation you have to shift the origin to (2,0) where t is on the x axis. This changes equation 1 tox = u(t-2) + 1/2 a (t-2)^2Now you can easily compare the two equations.Hope this helped.