## anonymous one year ago At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at t = 5.0 s? The acceleration-time graph looks like a straight line and the slope is negative and it makes a triangle with base of 6 units and a height of 6 units.

1. IrishBoy123

|dw:1444427089918:dw|

2. IrishBoy123

if that drawing is correct, you have $$a(t) = 6-t$$ you know that velocity $$v(t) = \int a(t) \; dt \; [ [A]]$$ you know that displacement/position is given by $$x(t) = \int \; v(t) \; dt\; [ [B] ]$$ you also have boundary conditions: $$x(0) = 25 m, v(0) = +15 m/s$$ integrate [A] then [B], and then use the conditions to solve for the constants then solve for x(5).

3. anonymous

Sorry for taking so long to replay but the site was down for me for some reason. That worked perfectly i got the equation from the graph and integrated it twice like you said and i got the correct answer, so thank you!