Monday

December 22, 2014

December 22, 2014

Posted by **Anonymous** on Saturday, July 7, 2012 at 10:51pm.

s(t)=-2t^3+15t^2-24t-6 [0,5]

- calculus -
**Damon**, Saturday, July 7, 2012 at 11:41pmv = ds/dt = -6 t^2 + 30 t -24

a = dv/dt = -12 t + 30

at t = 0 the point is at -6

at t = 1 it is at -17

at t = 2 it is at -10

etc until

at t = 5 it is at -1

graph that

the velocity changed sign in there, look for where v = 0

-6 t^2 +30 t -24 = 0

-t^2 + 5 t - 4 = 0

t^2 -5 t + 4 = 0

(t -4)(t-1) = 0

at t = 1 and at t = 4 the direction reverses

at t = 0 the velocity is -24 so it starts out at -6 moving south fast

at t = 1 it reverses course and starts up from -17 etc

at t = 4 it is at 10 and starts down toward reaching 1 at t = -5

**Answer this Question**

**Related Questions**

Calculus - Give acceleration a=d^2s/dt^2,initial velocity, and initial position ...

Calculus - Give acceleration a=d^2s/dt^2,initial velocity, and initial position ...

Calculus - Mathematica Project - Position, Velocity, and Acceleration? A rocket ...

calculus - Given the position function of an object moving in a straight line, ...

Math: Calculus - The velocity function is v(t)= t^2-5t+ 4 for a particle moving ...

calculus - the position of a particle moving along a coordinate line is s=√...

calculus - Find the equation of the line tangent to the graph of the given ...

Calculus - The motion of a spring that is subject to dampening (such as a car's ...

calculus - Consider the following. y = x2 − 9x x2 + 5x at (3, − 3 4...

Physics - A particle is moving along a straight line and its position is given ...