An object's velocity in m/s is given by the equation vx(t) = 10t-2. if it starts at x=0 at t=0, what is its position after 5 seconds?

d(5) = 115m

Position after 5 seconds is 115m

Explanation:

Given;

Initial position d(0) = 0

Time = 5 sec

Velocity function vx(t) = 10t - 2

To determine its position after 5 sec we need to calculate the position function.

d(t) = integral of vx(t)

d(t) = ∫10t - 2

d(t) = (10/2)t^2 - 2t + c

d(t) = 5t^2 - 2t + d(0)

c = d(0) = 0

d(t) = 5t^2 - 2t

So, at time t = 5

d(5) = 5(5^2) -2(5) = 125 - 10

d(5) = 115m

i dont have a visual in front of me, but i can you reason to try to get the right one. the graph should look like this, 0 to 6 there is a spike, 12 to 14 there should be a very low point. slight increase in the graph at 12 to 14, and flat 14 to 18

explanation: