[O0OITSTIFFANYY]

I'm really bad at science, so please help me answer these questions! And if its not a bother please show me how you did it..

Equations:
v = d/t
a = [delta]v/t
Vf = Vi + at
d= Vit+1/2at^2
Vf^2 = Vi^2 + 2ad

v = average velocity
d = distance
t = time
a = acceleration
Vf = final velocity
Vi = initial velocity



Consider the plot below describing motion
along a straight line with an initial position of
x0 = 10 m.


What is the position at 2 seconds? Answer in
units of m.

What is the position at 6 seconds? Answer in
units of m.

What is the position at 9 seconds? Answer in
units of m.




I REALLY APPRECIATE THE HELP! THANK YOUU!!

[xingjing]

The key to this is understanding that as you integrate velocity with respect to time, you obtain a change in displacement. If you haven't taken calculus, don't worry - basically what I'm saying is that when you take the area under the graph of velocity vs. time, you obtain a net change in displacement.

Take part a for example. The area under the graph is a triangle, whose area is 1/2 * base * height or 1/2 * 2* 9 = 9. Thus the net change in displacement is +9. And since x= x0 + vt, then x = 10 + 9 (the net change in displacement you found from the graph, which is much like velocity times time). Thus, you should obtain an answer of x = 19 at t = 2.

Do the same for the other two parts. For part c, be careful - since the graph goes under the x-axis at time 6, just take the area between the graph and the x-axis as the area under the graph.

Was that clear? Sorry if it wasn't.

[O0OITSTIFFANYY]

The key to this is understanding that as you integrate velocity with respect to time, you obtain a change in displacement. If you haven't taken calculus, don't worry - basically what I'm saying is that when you take the area under the graph of velocity vs. time, you obtain a net change in displacement.

Take part a for example. The area under the graph is a triangle, whose area is 1/2 * base * height or 1/2 * 2* 9 = 9. Thus the net change in displacement is +9. And since x= x0 + vt, then x = 10 + 9 (the net change in displacement you found from the graph, which is much like velocity times time). Thus, you should obtain an answer of x = 19 at t = 2.

Do the same for the other two parts. For part c, be careful - since the graph goes under the x-axis at time 6, just take the area between the graph and the x-axis as the area under the graph.

Was that clear? Sorry if it wasn't.

Thanks! I got the first two parts down but I still don't understand the third ..
Can someone help clarify how you do the third part?

[joie.de.vivre]

you subtract that triangular area from t=6 'til t=9 from the area under the graph from t=0 'til t=6.
because position is the antiderivative [you've learned calculus, yeah? it doesn't matter that much, anyway, here] of velocity

[O0OITSTIFFANYY]

you subtract that triangular area from t=6 'til t=9 from the area under the graph from t=0 'til t=6.
because position is the antiderivative [you've learned calculus, yeah? it doesn't matter that much, anyway, here] of velocity

Is the answer 57?
I tried putting that answer, but the computer said it was wrong

[joie.de.vivre]

um is it 45 for c then?
(35+10)

[xtah_]

hint :
area under graph = distance travelled; do remember to +10m (initial position)
negative velocity = negative direction (meaning it gets closer to the initial position[t=0s] if its on the positive side)