In the previous problem, the solution of the initial value problem is
A) $x = 16 t ^ { 2 } + 40 t + 200$
B) $x = - 16 t ^ { 2 } + 200 t + 40$
C) $x = - 16 t ^ { 2 } + 40 t + 200$
D) $x = 32 t ^ { 2 } + 40 t + 200$
E) $x = - 32 t ^ { 2 } + 40 t + 200$

Question

Quizplus · Accepted Answer

The correct solution to an initial value problem involving acceleration, velocity, and position typically involves integrating the acceleration to get velocity and then integrating velocity to get position. The constants of integration are determined by the initial conditions. The correct choice, $x = - 16 t ^ { 2 } + 40 t + 200$, likely represents a scenario where the acceleration is constant (due to the $t^2$ term), and the initial velocity and position are represented by the linear term and the constant term, respectively. This choice correctly integrates acceleration (implied to be constant and negative, indicating a downward force such as gravity) to get velocity and then position, with appropriate initial conditions.

In the Previous Problem, the Solution of the Initial Value x=16t2+40t+200x = 16 t ^ { 2 } + 40 t + 200x=16t2+40t+200

In the Previous Problem, the Solution of the Initial Value $x = 16 t ^ { 2 } + 40 t + 200$