What distance will be covered before coming to stop?

[chwala]

Homework Statement

A train braking with constant deceleration covers $1$ km in $20s$, and a second kilometre in $30s$. Find the deceleration. What further distance will it cover before coming to a stop, and how long will this take?

Relevant Equations

Mechanics

Phew; back n forth on this anyway, my lines

...

$v_{1} = \dfrac{1000}{20} = 50$ m/s in the first $10$ seconds.


$v_{2} = \dfrac{1000}{30} = \dfrac{100}{3}$m/s in the first $35$ seconds.

where $v_1$ and $v_2$ are the respective velocities in $10s$ and $35s$ respectively.

Thus,

$a = \dfrac{33.3333 - 50}{35-10} = -\dfrac{2}{3}m/s^2$.

Therefore deceleration is equal to $\dfrac{2}{3} m/s^2= 0.666m/s^2$.

For next part,

$v=u+at$

$v_{3}= 33.333+(-0.666×15)$

$v_{3} = 23.3333$ m/s

using $v^2 =u^2 +2as$

$0 = 23.333 + (2×-0.666×s)$

$544.3 = 1.333s$

$s = 408.3m$ correct to one decimal place.

Finally,

Using $v= u+at$

$0 = 23.333+ (-0.666t)$

$23.333= 0.66t$

$t =35$ seconds

There could be a much better approach hence my post. Cheers.

 

[Vanadium 50]

There could be a much better approach hence my post.

There is. I can't read the problem, but I can tell you that plugging in numbers is the last step, not the first.

 

[chwala]

There is. I can't read the problem, but I can tell you that plugging in numbers is the last step, not the first.

I re-typed the question.

 

[jbriggs444]

$v_{1} = \dfrac{1000}{20} = 50$ m/s in the first $10$ seconds.

I agree.

$v_{2} = \dfrac{1000}{30} = \dfrac{100}{3}$m/s in the first $35$ seconds.

Where did 35 come from?
Why do you care about the average velocity for the first 35 seconds?
Did you actually compute the average velocity for the first 35 seconds?
Out of purely idle curiosity, what would be the average velocity over the second interval?

 

[chwala]

I agree.

Where did 35 come from?
Why do you care about the average velocity for the first 35 seconds?
Did you actually compute the average velocity for the first 35 seconds?
Out of purely idle curiosity, what would be the average velocity over the second interval?

For an object moving with constant acceleration or deceleration over a period of time, then it is common knowledge that these quantities are the same;

• average velocity
• velocity when half the time has passed
• mean of the initial and final velocities

Where did 35 come from?

From $t = 20 + \dfrac{30}{2} s$.


Did you actually compute the average velocity for the first 35 seconds?

In the last $30$ seconds, the average velocity is $\dfrac{1000}{30} m/s$, so this is the velocity after $(20 +15) = 35$ seconds, from this i am able to determine the rate of deceleration as indicated.

 

[jbriggs444]

In the last $30$ seconds, the average velocity is $\dfrac{1000}{30} m/s$, so this is the velocity after $(20 +15) = 35$ seconds, from this i am able to determine the rate of deceleration as indicated.

Ahhh, there seems to be a language difficulty.

You'd originally written:

$v_{1} = \dfrac{1000}{20} = 50$ m/s in the first $10$ seconds.

This is phrased unfortunately. I would have said instead:

The average velocity is 50 m/s over the first 20 seconds.

Given constant acceleration, it follows that the instantaneous velocity is 50 m/s at the 10 second mark".

Similarly:

$v_{2} = \dfrac{1000}{30} = \dfrac{100}{3}$m/s in the first $35$ seconds.

Where I would have said:

The average velocity is $\frac{1000}{30} \text{ m/s}$ over the next 30 seconds.

Given constant acceleration, it follows that the instantaneous velocity is $\frac{1000}{30} \text{ m/s}$ at the 35 second mark.

 

[chwala]

Ahhh, there seems to be a language difficulty.

You'd originally written:

This is phrased unfortunately. I would have said instead:

The average velocity is 50 m/s over the first 20 seconds.

Given constant acceleration, it follows that the instantaneous velocity is 50 m/s at the 10 second mark".

Similarly:

Where I would have said:

The average velocity is $\frac{1000}{30} \text{ m/s}$ over the next 30 seconds.

Given constant acceleration, it follows that the instantaneous velocity is $\frac{1000}{30} \text{ m/s}$ at the 35 second mark.

Were you having a different approach where $35$ is not going to be considered as part of working?

 

[jbriggs444]

Were you having a different approach where $35$ is not going to be considered as part of working?

No. I was merely concerned that an average velocity over a period of 35 seconds would have nothing to do with the solution.

Your original statements appeared to claim an average velocity of $\frac{1000}{30} \text{ m/s}$ over the interval from $t=0$ to $t=35$.

Yes, the computations you performed match the ones would have performed. The concern was with the explanations [or lack thereof] for those computations.

The advice from @Vanadium 50 in #2 is apt: Explanations first, computations after.

 

[kuruman]

The formal approach recognizes that one is given two pairs of positions and clock times. Assuming that $t=0$ is when the train starts braking, one has
$(x_1,t_1)=(1~{\text{km}},20~{\text{s}})$
$(x_2,t_2)=(2~{\text{km}},50~{\text{s}})$

Then one writes the position SUVAT equation twice to get a system of two equations and two unknowns
$x_1=v_0t_1+\frac{1}{2}at_1^2$
$x_2=v_0t_2+\frac{1}{2}at_2^2$
The unknowns are the initial velocity $v_0$ and acceleration $a$.

I leave the rest as an exercise to the student.
Hint: Divide the first equation by $t_1$, the second equation by $t_2$ and subtract.

Substitute numbers at the very end.

 

[chwala]

The formal approach recognizes that one is given two pairs of positions and clock times. Assuming that $t=0$ is when the train starts braking, one has
$(x_1,t_1)=(1~{\text{km}},20~{\text{s}})$
$(x_2,t_2)=(2~{\text{km}},50~{\text{s}})$

Then one writes the position SUVAT equation twice to get a system of two equations and two unknowns
$x_1=v_0t_1+\frac{1}{2}at_1^2$
$x_2=v_0t_2+\frac{1}{2}at_2^2$
The unknowns are the initial velocity $v_0$ and acceleration $a$.

I leave the rest as an exercise to the student.
Hint: Divide the first equation by $t_1$, the second equation by $t_2$ and subtract.

Substitute numbers at the very end.

...
I was able to realize,

$2000 = 40v_0 +400a$ and
$4000 = 100v_0 +2500a$

Solving for the unknown, i can see that $a= -\dfrac{2}{3} m/s^2$

Now to my next question, for the unknown $v_0$ i have $v_0 = 56.7m/s$. This is the initial velocity at what point? before deceleration?

 

[haruspex]

...
for the unknown $v_0$ i have $v_0 = 56.7m/s$. This is the initial velocity at what point? before deceleration?

Yes.

For an ad hoc approach, it slowed from an average of 50m/s over 20s to 33.33m/s over 30s. Those average speeds would have been the instantaneous speeds in the centres of those intervals, so it slowed from 50 to 33.33 in 20/2+30/2=25 seconds.

 

[kuruman]

...
I was able to realize,

$2000 = 40v_0 +400a$ and
$4000 = 100v_0 +2500a$

Solving for the unknown, i can see that $a= -\dfrac{2}{3} m/s^2$

Now to my next question, for the unknown $v_0$ i have $v_0 = 56.7m/s$. This is the initial velocity at what point? before deceleration?

You missed my point in post #9. The idea is not to susbstitute numbers in the middle of the solution. Here is what I had in mind.
Divide each equation by the times
$$\begin{align} & x_2/t_2=v_0+\frac{1}{2}at_2 \nonumber \\& x_1/t_1=v_0+\frac{1}{2}a t_1 \nonumber \end{align}$$Subtract bottom equation from top and solve for the acceleration $$\begin{align}
& x_2/t_2-x_1/t_1=\frac{1}{2}a(t_2 -t_1) \nonumber \\
&a=\frac{ x_2/t_2-x_1/t_1}{\frac{(t_2 -t_1)}{2}}. \nonumber
\end{align}$$This symbolic expression shows what's going on here. The last equation is a straightforward application of $a=\Delta v/\Delta t$ where $v$ is the instantaneous velocity. Note that the average velocity over interval $t_2$ is $\bar v_2=x_2/t_2$ and is equal to the instantaneous velocity $v_2$ at time $t'_2=\frac{1}{2}t_2$. Likewise, the average velocity over interval $t_1$ is $\bar v_1=x_1/t_1$ and is equal to the instantaneous velocity $v_1$ at time $t'_1=\frac{1}{2}t_1$. Thus the acceleration is $$a=\frac{v_2-v_1}{t'_2-t'_1}=\frac{ x_2/t_2-x_1/t_1}{\frac{(t_2 -t_1)}{2}}.$$My point is the same as @Vanadium 50 raised. You don't get a deeper understanding of what's going unless you substitute numbers at the very end. You could have written the expression for the acceleration immediately by putting together these three ideas when an object moves under constant acceleration

1. The average velocity is the position change over the time interval of the change.
2. The instantaneous velocity is equal to the average velocity at half the time interval of the change.
3. The acceleration is equal to the change of instantaneous velocity over the time interval of the change.

 

[kuruman]

This is the initial velocity at what point? before deceleration?

What do you think? What does the expression for the instantaneous velocity look like?