At time t = 0 a particle leaves the origin and moves along the x-axis. At time t seconds, the velocity of P is v m/s in the positive x direction, where v=4t^2–13t+2. How far does it travel between the times t1 and t2 at which it is at rest?

First, you have to work out the values of t1 and t2 at which the particle is at rest. This is done by solving the quadratic equation for v, producing values for t of 13/8 +- sqrt(137): 0.1619s and 3.088s. To calculate the distance travelled between the points, you simply integrate the equation for v to find x, between t1 and t2. This gives x=(4/3)t^3–(13/2)t^2+2t+C, where the integration constant C is 0 as it starts at the origin (this is also irrelevant as it is a definite integral!) When the values for t1 and t2 are placed in the equation, it produces a value for distance between the times of 137*sqrt(137)/96, approximately equal to 16.704.