Question

Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units of Complete parts a through c. a. Determine when the motion is in the positive direction and when it is in the negative direction. b. Find the displacement over the given interval. c. Find the distance traveled over the given interval. vt)-32 - 24t+36; [0.7]

Answer 1

Answer:

a) positive direction: t < 2s & t>6s   ; negative direction: 2s < t < 6s

b) 7 m

c) 71 m

Step-by-step explanation:

Given:

v(t) = 3t^2 -24t +36   [0 , 7]

Find:

a) The value of time when particle is moving in positive direction:

The change in direction of the particle can be determined by v(t) > 0

Hence,

                                    0 < 3t^2 -24t +36

                                    0 < t^2 - 8t + 12

                                    0 < (t - 2)*(t - 6)

                                    t < 2s  , t  > 6s  

The particle travels in positive direction in the interval t < 2s and t > 6s , While it travels in negative direction when 2s < t < 6s.

b) The displacement ds over the given interval [ 0 , 7 ]

                                  ds = integral (v(t)).dt

                                  ds = t^3 -12t^2 +36t

                                  ds = 7^3 -12*7^2 +36*7

                                  ds = 7 m

c) Total distance traveled in the interval:

                                 Total distance= ds(0-2) + ds(2-6) + ds(6-7)

                                 D = 2*(2^3 -12*2^2 +36*2) - 2*(6^3 -12*6^2 +36*6) + 7

                                 D = 2*32 - 2*0 + 7

                                 D = 71 m