Asked by Shelly
A car starts 200 m west of the town square and moves with a constant velocity of 15 m/s toward the east.
a.Write the equation that represents the motion of the car.
b.Where will the car be 10 minutes later?
c.When will the car reach the town square?
a.Write the equation that represents the motion of the car.
b.Where will the car be 10 minutes later?
c.When will the car reach the town square?
Answers
Answered by
Damon
if the town square is at x = 0 and east is positive then:
x = -200 + 15 t
in ten minutes x = -200 + 150 = -50 or 50 meters west of the square
when is x = 0?
0 = -200 + 15 t
t = 200/15 = 13 1/3 seconds to the square
x = -200 + 15 t
in ten minutes x = -200 + 150 = -50 or 50 meters west of the square
when is x = 0?
0 = -200 + 15 t
t = 200/15 = 13 1/3 seconds to the square
Answered by
Kat
Ok so I’m not going to do the whole question over again but here’s what I have:
(200m) / (15m/s) = 13.33333 sec
As far as where the car is after 10mins, my ANSWER KEY says *8800m* but I seem to keep getting 9000m.
Ex. (Vi + Vf) t
/2
[excuse my inability to format this, and I’m too lazy to put in units]
First, 10min = 600sec (t)
(15+15)= 30
30/2 = 15
15 x 600 = 9000m
But whatever lol. I hope this helped in some way.
(200m) / (15m/s) = 13.33333 sec
As far as where the car is after 10mins, my ANSWER KEY says *8800m* but I seem to keep getting 9000m.
Ex. (Vi + Vf) t
/2
[excuse my inability to format this, and I’m too lazy to put in units]
First, 10min = 600sec (t)
(15+15)= 30
30/2 = 15
15 x 600 = 9000m
But whatever lol. I hope this helped in some way.
Answered by
V
I just wanted to add onto what Kat said- I also kept getting 9000, but since you start at -200m, I subtracted 200m from the 9000m to get the 8800m. I think that's how you're supposed to do it. :)
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