# Problem: At t = 0, an object of mass m is at rest at x = 0 on a horizontal, frictionless surface. A horizontal force Fx = F0 (1 − t/T), which decreases from F0 at t = 0 to zero at t = T, is exerted on the object.(a) Find an expression for the object's velocity at time T.(b) Find an expression for the object's position at time T.

###### FREE Expert Solution

This problem requires us to find the velocity and position expressions for an object at a specific time given a force function.

For a Force with Calculus type of problem, we follow these steps:

1. Use the Newton's Second Law equation ΣF=ma to find a function for the acceleration
2. Integrate the acceleration function to get velocity (don't forget about the integration constant!)
3. Integrate the velocity to get the position (if required)
4. Calculate the position and/or velocity at a specific time of interest (if required)

A diagram like this one can help you remember the relationships between the variables:

$\mathbit{P}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\underset{\frac{\mathbit{d}}{\mathbit{d}\mathbit{t}}}{\overset{{\mathbf{\int }}{\mathbit{d}}{\mathbit{t}}}{\mathbit{V}}}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\underset{\mathbit{F}\mathbf{=}\mathbit{m}\mathbit{a}}{\mathbit{A}\mathbf{,}\mathbit{F}}$

Remember the power rule of integration.

To integrate,

, where C is the constant of integration.

The velocity function is the integral of the acceleration function, a(t). We're not given a(t), but we are given a function for force, F(t).

The object's position function is the expression for x(t), which is the integral of the velocity function v(t).

So here's our process:

###### Problem Details

At t = 0, an object of mass m is at rest at x = 0 on a horizontal, frictionless surface. A horizontal force Fx = F0 (1 − t/T), which decreases from F0 at t = 0 to zero at t = T, is exerted on the object.

(a) Find an expression for the object's velocity at time T.

(b) Find an expression for the object's position at time T.