I am really looking for someone to guide me through this. If you just want to give me the answer that's fine, but I would much rather know how to do this one.

The equation for a toy car is given by x(t)=.5t-3.5. Find the position at 4 seconds. At what instant is the position zero for x(t) =.5t-3.5 and for x-4. Also, how long will it take for the position to reach 12 meters.

Question

I am really looking for someone to guide me through this. If you just want to give me the answer that's fine, but I would much rather know how to do this one.

The equation for a toy car is given by x(t)=.5t-3.5. Find the position at 4 seconds. At what instant is the position zero for x(t) =.5t-3.5 and for x-4. Also, how long will it take for the position to reach 12 meters.

anonymous · Answer

x(t) is the distance x of time

x(t) = .5t -3.5.
therefore, at time t = 4 seconds

x(4) = .5(4) - 3.5         =    -1.5  meters


at what point is the car at position 0 or

x(t) = 0  =  .5t - 3.5
.5t = 3.5 
t = 7 seconds

anonymous · Answer

Same equations. That makes no sense. I was sure this LAB wants me to use different equations. Maybe its tricking me.

anonymous · Answer

or maybe I don't know what I am talking about, thanks man..lol

anonymous · Answer

nm I know what I am doing wrong!

anonymous · Answer

verify this for me if you look at this again! y=mx+b therefore  m=velocity and b= position. therefore, original equation is  x(t)=.5(t)-3.5. If I am creating an equation for 4 seconds, then the 3.5 would become 4 in the original equation correct?

anonymous · Answer

so the next equation would be x(t)-.5(t)-4. Then evaluate zero at t=4?

anonymous · Answer

err = not -