A particle moves along the x axis. Its position is given by the equation
x = 1.9 + 2.5t − 3.7t2 with x in meters and t in seconds.

(a) Determine its position when it changes direction.
____m

(b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.)
____m/s

Question

A particle moves along the x axis. Its position is given by the equation 
x = 1.9 + 2.5t − 3.7t2 with x in meters and t in seconds.

(a) Determine its position when it changes direction.
 ____m

(b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.)
 ____m/s

anonymous · Answer

$$x=1.9+2.5t+3.7t ^{2}$$
$$V=2.5-3.7t$$

V=0 when it stops 
0=2.5-7.4t
t=0.3378

a) $$x=1.9+2.5t+3.7t ^{2}$$  (t=0.3378)

x=3.1167m

b) $$V=2.5-3.7t$$
(t=0) 
V=2.5  but because it's returning here velocity is -2.5

anonymous · Answer

the equations must be $$x=1.9+2.5t-3.7t ^{2}$$

all of the others are correct

anonymous · Answer

both answers are wrong...

anonymous · Answer

lol, what's the correct ones ?