Calculating Distance Covered by a Particle Using Displacement Vector

[Nader AbdlGhani]

Homework Statement

The displacement vector of a particle of mass 50 gm. is given as a function in time (t) by the relation $\vec S$=(5t-t²) $\hat c$ where $\hat c$ is a constant unit vector , s is measured in centimetre and t in second .Knowing that the particle started its motion at t=0
Find :
(1) The force vector acting on the particle and the work done by this force during the first five seconds of motion.
(2) The total distance covered during the first three seconds of motion.

Homework Equations

$w=\vec F \cdot \vec S$
$F=ma$

The Attempt at a Solution

The requirement number (1) is quite easy by differentiating vector s with respect to time 2 times we get the acceleration vector and we already have the mass , so we can get the force easily, but the problem is in getting the total distance covered , in my best attempt , I could only get the maximum displacement by differentiating vector s with respect to time and equal it with zero we will get 5-2t=0 so displacement is maximum when t=2.5 seconds , what should I do next ?

 

[Buzz Bloom]

Hi Nadar:

What is the displacement at t=0?
What is the displacement at t=3?
How far did it move between t=0 and t=3?

Hope this helps.

Regards,
Buzz

 

[gneill]

I think you want to investigate the arc length of a curve. The curve in this case is one dimensional (lies along a straight line), can you think of a way to sum up the distance over time?

 

[Nader AbdlGhani]

Hi Nadar:

What is the displacement at t=0?
What is the displacement at t=3?
How far did it move between t=0 and t=3?

Hope this helps.

Regards,
Buzz

Thanks , The problem is with the distance not the displacement .

 

[Nader AbdlGhani]

I think you want to investigate the arc length of a curve. The curve in this case is one dimensional (lies along a straight line), can you think of a way to sum up the distance over time?

Nope .