Kinematics question please help

Question

A particle moves on the x axis. It starts from rest at the origin O and moves in the positive direction with constant acceleration of 2 m/s^2 until its velocity reaches 10 m/s. It then continues with constant velocity of 10 m/s for some time, before decelerating uniformly and coming to rest after a total time of 20 s. The total distance travelled is 160 m. Find the magnitude of the deceleration.

Answer 1

Hi Orawan:

 

   First, the fact that kinematic equations only apply while acceleration is constant means we have to look at motion like this in terms of intervals.  There are three intervals over which there is constant acceleration; we can call them the first, second, and third leg.  Ultimately, we want to determine the acceleration in the last leg.

 

This problem has mixed around what we know a little, so we have to sort that out.  But a good place to start to organize the given information is always to start by attacking directly what you want to solve for, which means working on the third leg.  In that leg, we have:

 

to find: a

know:  v_o, v

 

We know v_o = 10 m/s, because that was the speed at the end of the first leg, and throughout the entire (constant velocity) second leg, and thus at the start of the third leg.

 

And we know v = 0, because it comes to rest.

 

But we do not know a third quantity for the third leg (either t or (x-x_o)), so we are stuck trying to solve for a in the third leg with just that information (we need three known kinematic quantities).  We expected this, but going at this systematically helps organize the information.

 

Now we know we have to solve for either the displacement or the time of the third leg.  We are given time and displacement values, but only for all three legs together.  So we have to start breaking them down leg by leg, to extract the information we need for just the third leg.

 

Let's look at the first leg, since we have the most specific information about that.  Let's try to solve for displacement first (we could also solve for time first; there is no particular advantage to solving for one or the other first).

 

to find: (x-x_o)

know:  v_o, v, a

 

v_o = 0 (starts at rest)

v = 10 m/s

a = 2 m/s²

 

Choose the kinematic equation that includes these four quantities:

 

v² = v_o²+2a(x-x_o)

 

We can solve for displacement on the first leg with this equation.  I obtained (x-x_o)₁ = 25 m

 

To solve for time, we can use the same knowns, and the kinematic equation involving those knowns and time.

 

v = v_o+at  -->  I get t₁ = 5 s

 

So we now know the part of displacement and time taken up by the first leg.  This means the total displacement and total time for the last two legs must be 135 m and 15 s.

 

For the second leg, we know a = 0, so the only kinematic relationship in play is the one for constant velocity: (x-x_o) = vt, with v = 10 m/s.  However, displacement and time are still unknown, just as they are for the third leg.

 

Where this is leading is we are going to have to set up a system of multiple equations with multiple unknowns.

 

(x-x_o)₂ = 10t₂  for the second leg

 

(x-x_o)₃ = (1/2)(v+v_o)t₃ = (1/2)(10+0)t₃ = 5t₃  for the third leg

 

The reason I set up these equations involving the displacements and times for the second and third leg is because we have other relationships between those quantities, based on what was calculated above:

 

(x-x_o)₂ + (x-x_o)₃ = 135   (the total displacement of legs 2 and 3 is (+)135 m)

t₂+t₃ = 15  (the total time for legs 2 and 3 is 15 s)

 

This means we can convert one of the two equations into an equation involving only the unknowns from the other equation, since:

 

(x-x_o)₂ = 135 - (x-xo)₃

t₂ = 15 - t₃

 

Now I can convert the second-leg equation into one relating (x-x_o)₃ and t₃:

 

(x-x_o)₂ = 10t₂ -->   135-(x-x_o)₃ = 10(15-t₃)

 

I am now going to rearrange this equation to write t₃ in terms of (x-x_o)₃:

 

t₃ = 1.5 + (1/10)(x-x_o)₃

 

Putting this into third-leg relationship above between (x-x_o)₃ and t₃ gives:

 

(x-x_o)₃ = 5t₃ --> (x-x_o)₃ = 5(1.5 + (1/10)(x-x_o)₃)

 

Solving this for (x-x_o)₃ , I obtained (x-x_o)₃ = 15 m

 

This is now the third piece of information we need to solve for a on the third leg... but I still went back and solved for all of t₃, (x-x_o)₂, and t₂, just to make sure the results were consistent with what was given (they were; I got t₃ = 3 s, t₂ = 12 s, and (x-x_o)₂ = 120 m; this gave me confidence going on to solve for a in the third leg).

 

Now, for the third leg, we have:

 

to find:  a

know:  v, v_o, (x-x_o)₃

 

Put them together in the appropriate kinematic equation for those four quantities, and solve for a.  I obtained:

a₃ = -3.33 m/s²  (negative, as would be expected from slowing down when the object is moving in the positive direction)

 

As you have seen, I can make a calculation error as well as anyone, so no absolute guarantees on this result.  But the procedure is sound.  If you have any questions, just let me know.