The acceleration of a particle is given by a(t)= -2.00 m/s^2 + (3 m/s^3)t. Required:a. Find the initial velocity vo such that the particle will have the same x-coordinate at t=4.00 s as it had at t=0. b. What will be the velocity at t=4.00 s ?

Explanation:

a(t)= -2.00 m/s^2 + (3 m/s^3)t.

dv / dt =  -2.00 m/s^2 + (3 m/s^3)t.

dv = (-2.00 m/s^2 + (3 m/s^3)t.)dt

v  = - 2t + 3 t² / 2 + c , where c is a constant

for initial velocity t = 0

v0 = c

v  = - 2t + 3 t² / 2 + v0

ds / dt =  - 2t + 3 t² / 2 + v0

ds =  (- 2t + 3 t² / 2 + v0)dt

s = - 2t²/2 + 3 t³/6 + vot + c₁

At t = 0

s = c₁

At t = 4

s = -16 + 32 + 4v0 + c₁

=  4v0 + c₁ + 16

Given

4v0 + c₁ + 16  = c₁

v0 = - 4 m /s

Putting this value in the equation of velocity

v  = - 2t + 3 t² / 2 - 4

At t = 4

v = -8 + 24 - 4

= 12 m / s

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .