A mysterious rocket-propelled object of mass..A mysterious rocket-propelled object of mass 45.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered...

A mysterious rocket-propelled object of mass..

A mysterious rocket-propelled object of mass

45.0 kg is initially at rest in the middle of the horizontal, frictionless

surface of an ice-covered lake. Then a force directed east and with

magnitude ` F(t) = (16.8N/s)t` is applied. How far does the object

travel in the first 5.00 s after the force is applied?

txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

Let's consider Newton's second law to solve this problem:

` ` `F = ma`

Now, let's substitute our values for mass and force:

`16.8 t = 45.0*a`

Solving for acceleration (and simplifying the fraction):

`a = 28/75t `

Now, let's consider that acceleration is a function of time. Also, remember that velocity will be the antiderivative of acceleration. So, let's solve for velocity:

`v(t) = int_0^t 28/75tau d tau`

Solving the integral:

`v(t) = 28/75 t^2/2 + v_0 = 14/75 t^2 + v_0`

To continue, we'll need to find that initial velocity. Thankfully the problem states we started at rest! This phrasing implies that our initial velocity will be zero giving us our final velocity function:

`v(t) = 14/75 t^2`

Now, we can continue by knowing that displacement is the antiderivative of velocity:

`s(t) = int_0^t v(tau) d tau = int_0^t 14/75 tau^2 d tau`

Solving the integral:

`s(t) = 14/75 t^3/3 + s_0`

Simplifying:

`s(t) = 14/225 t^3 + s_0`

Now, to find the initial displacement, we just set the start point however we want! It does not matter one bit where we put our axes for this part. So, let's just say that our rocket-propelled object starts at a displacement of 0! This value gives us the following function:

`s(t) = 14/225 t^3`

Now, all we need to do is substitute the given value for time into the equation:

`s(5.00) = 14/225*5.00^3 = 14/225*125.00 = 70/9 = 7.78`

So, the distance travelled will be 7.78 meters.

Just remember how force relates to acceleration through Newton's Second Law, then just integrate to find velocity and displacement. That method or some slight variation of it is generally how these problems are solved!

I hope that helps!

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