Question

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.

Answer 1

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).

Answer 2

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!