The position of a particle P on a line l is given by s(t) = t^2 − 4t, where s(t) is in feet and t is in seconds. Find the velocity as a function of time.

Question

v(t) = 2t-4

v(0) = -4

v(8)=12

Determine the time intervals (in seconds) in which the particle moves in the positive direction and the negative direction. (Enter your answers using interval notation.)

Answer 1

s(t) = t^2 -4t

v(t) = s'(t) = 2t-4 >0 when t>2

the object is moving in a positive direction in the time interval (2, infinity) when v(t)>0

v(t) = 2t-4>0

2t>4

t>4/2 =2

v(-4) = 2(-4)-4 =-8-4=-12

v(8) =2(8)-4= 16-4 = 12

t^2-4t is graphically an upward opening parabola with x intercepts x=0 and x=4

axis of symmetry is x=2, vertex is (2,s(2)) = (2, -4)

to the right of the vertex the parabola is upward sloping. the slope = the velocity, It's >0 to the right of the vertex, to the right of x=2 where x>2