An object moves such that acceleration a(t)=2(1-3t). Given v(0)=-5 and x(0)=4, find x(1). I know the answer is -1, just not too sure on how to get the work, thanks!

x(1) = -1

Step-by-step explanation:

Given the equation for acceleration of the object as;

a(t)=2(1-3t)

a(t) = 2-6t

Integrate to get the velocity

v(t) = 2t-6t²/2 + C

v(t) = 2t-3t²+C

If v(0) = -5

when t = 0 v = -5

-5 = 2(0)-3(0)²+C

-5 = C

C = -5

v(t) = 2t-3t²-5

To get the displacement x(t), we will integrate the velocity function

x(t) = 2t²/2-3t³/3-5t+C

x(t) = t²-t³-5t+C

If x(0) = 4

At t = 0, x(t) = 4

4 = 0²-0³-5(0)+C

4 = C

C = 4

x(t) = t²-t³-5t+4

Get x(1)

x(1) = 1²-1³-5(1)+4

x(1) = 1-1-5+4

x(1) = -5+4

x(1) = -1

Hence x(1) = -1

-60.75

step-by-step explanation:

just add the numbers then put the negative sign on it