Help with Calc problem: particle movement along x-axis involving acceleration and velocity! Can any1 help?

Question

Can any1 help me with this and sorta explain it a little bit too.

A particle moves along the x-axis so that its acceleration at any time t is given by a(t)=6t-18. At time t=0 the velocity of the particle is v(0)=24, and at time t=1 its position is x(1)=20.

(a) Write an expression for the velocity v(t) of the particle at any time t.

(b) For what values of t is the particle at rest?

(c) Write an expression for the position x(t) of the particle at any time t.

(d) Find the total distance traveled by the particle from t=1 to t=3.

PLEASE!! I AM BEGGING FOR HELP ON THIS!!!

Answer 1

Remember - acceleration is the time derivative of velocity, which is the time derivative of position. If a is acceleration, v is velocity and x is position, then

a = dv/dt = d^2x/dt^2

v = dx/dt

(a) Integrate a wrt t:

integ(6t - 18) dt = 3t^2 - 18t + C

A little work will show you that the C is the initial velocity, which you're given, so

v(t) = 3t^2 - 18t + 24

(b) You have a quadratic for v - find the roots to find where the particle is at rest, i.e. where v(t) = 0

(c) Integrate v wrt t to get x. This time, the constant you get will be the initial position, but you're given x(1). So plug 1 in for t, set it equal to 20, and solve for the constant of integration.

(d) This one may be tricky. To find the total distance travelled, you can't simply integrate, because negative distances will reduce the positive distances travelled. So, you'll have to find those points where the velocity goes to zero (aka part b). If any of those fall into the region [1, 3], you'll have to figure the distance travelled in each leg. For instance, if the velocity goes to 0 at t = 1.5, you'll have to find the distance travelled from 1 to 1.5 and the distance travelled from 1.5 to 3, then add them.

Good luck.

Answer 2

(a) take the integral of the acceleration equation to get the velocity equation: v(t)=3t^2-18t+C where C is a constant .... but since we know taht a t=0, v =24, substitue this is to find C ... V(0)=3(0)^2-18(t)+C=24 ... C=24 ... v(t)=3t^2-18t+24

(b)solve v(t) for zeros ... 3(t+4)(t-2) means that the zeros are a t=2 and t=-4

(c) take the integral of the velocity equation to get the position equation: x(t)=t^3-9t^2+24t+C where C is a constant .. since we know x(1)=20 substitue to find C ... x(1)=(1)^3-9(1)^2+24(1)+C=20 ... C=4 ... x(t)=t^3-9t^2+24t+4

(d) trick question ... since the particle changes direction at t=2 you have to find it's position at t=2 ... x(1)=20, x(2)=24, x(3)=22 ... so the particle moves from 20 to 24 and then back to 22 so the TOTAL distance is 6 ... :)

cheers

oh explain it too! ... here's how i remember acceleration/velocity/position problems.. think about throwing a ball through the air ... acceleration is constant (gravity), the velocity starts out really fast (i just threw it) and then slows down to nothing, and then speeds up as it falls back to the earth. finally the position kinda looks like a parabola. where's the calculus??? if the acceleration is constant (flat line ... at -9,81 m/s^2), the velocity is a straight line (with negative slope of -9.81), and the position is a parabola with a function like -9.82x^2+Ax+B ... if you remember that the acceleration is constant you can see that the velocity must be the integral of the acceleration, and also that the position is the integral of the velocity. horizontal line --> sloped line --> parabola where --> = integral

Answer 3

http://www.mecca.org/~halfacre/MATH/netchange.htm

This site has the same problem it may help,