I solved few questions on kinematics -- Is my working correct?

[Aichuk]

So the first question is:

A student standing on the platform of a railway station notices the first two carriages of an arriving train pass her in 2.0 s and the next two in 2.4 s. The train is decelerating uniformly. Each carriage is 20 m long. When the train stops, the student is opposite the last carriage. How many carriages are there?

My working went:

Avg. Velocity for first two carriages (Va) = 40 m/ 2.0 s = 20 m/s
Avg. Velocity for first two carriages (Vb) = 40 m/ 2.4 s = 16.7 m/s (correct to 3 significant figures).

Time at Va = 2.0 s / 2 = 1s

Time at Vb = 2 s + (2.4 s / 2) = 3.2 s

Acc. = (16.7 - 20 ) m/s / 2.2s = -1.5 ms^-2

At time = 1 s

Va = u + at
20 m/s = u - 1.5 m/s
Therefore, u = 21.5 m/s

Therefore, total time taken is

0 m/s = 21.5 m/s + (-1.5)(t)
t = -21.5/-1.5 = 14.3 s

So total distance is

s = ut + 1/2 at^2
= (21.5 m/s) (14.3s) + (-1.5 ms^-2) (1/2) (14.3^2)
=154 (approx.)

So by dividing the distance by the number of carriages,
154/20 = 7.7
approximated to 8 carriages

It matches with the answer in my book but is the way I solved the answer correct?

Also the next question is more about a concept. So check whether my answer is correct on this too..

In a manual, it's suggested that when driving at 13 m/s, a driver should always keep a minimum of two cars-lengths between the driver's car and the one in front.

Suggest a scientific justification, making reasonable assumptions for magnitudes of any quantities.

So my answer for (a) is that since that a car has a large mass and is moving relatively fast, so it has a high inertia. Thus it takes a large force for it to decelerate quickly so the minimum of two cars-length will allow it to steadily decelerate and come to rest. Any smaller space may cause the car to have to decelerate so quickly that the driver may be injured. Or it may not allow the car to decelerate and come to rest, causing it to hit the car in front.

I'm really doubtful of my answer to the second question especially my phrasing of the answer. Sorry for the long post, I just need some help. Thanks :)

 

[PeroK]

So the first question is:

A student standing on the platform of a railway station notices the first two carriages of an arriving train pass her in 2.0 s and the next two in 2.4 s. The train is decelerating uniformly. Each carriage is 20 m long. When the train stops, the student is opposite the last carriage. How many carriages are there?

My working went:

Avg. Velocity for first two carriages (Va) = 40 m/ 2.0 s = 20 m/s
Avg. Velocity for first two carriages (Vb) = 40 m/ 2.4 s = 16.7 m/s (correct to 3 significant figures).

Time at Va = 2.0 s / 2 = 1s

Time at Vb = 2 s + (2.4 s / 2) = 3.2 s

Acc. = (16.7 - 20 ) m/s / 2.2s = -1.5 ms^-2

At time = 1 s

Va = u + at
20 m/s = u - 1.5 m/s
Therefore, u = 21.5 m/s

Therefore, total time taken is

0 m/s = 21.5 m/s + (-1.5)(t)
t = -21.5/-1.5 = 14.3 s

So total distance is

s = ut + 1/2 at^2
= (21.5 m/s) (14.3s) + (-1.5 ms^-2) (1/2) (14.3^2)
=154 (approx.)

So by dividing the distance by the number of carriages,
154/20 = 7.7
approximated to 8 carriages

It matches with the answer in my book but is the way I solved the answer correct?

I like the logical way you solved it. I'm not sure why you doubt yourself.

You might have polished it off a little quicker. Once you have $u$ and $t$, then (for constant deceleration from $u$ to $0$) you have:

$s = ut/2$

 

[Meian man]

hi can you please tell me what book you're referring to?
thanks :)

 

[PeroK]

hi can you please tell me what book you're referring to?
thanks :)

This thread is over five years old!

 

[lim1029]

This question is from the book Cambridge International AS & A Level Physics Student's Book 3rd edition END OF TOPIC QUESTIONS 1 in chapter 2

 

[Kareemoya]

Hey I didn’t quite get the “time at Va and Vb” step can you explain pls

 

[haruspex]

Hey I didn’t quite get the “time at Va and Vb” step can you explain pls

The train is decelerating uniformly, so it loses equal amounts of velocity in equal intervals of time.
In particular, it loses the same amount in the first second as in the second second. So after the first second it has slowed to the average of its speed at the start and its speed after two seconds.

 

[kuruman]

It should also be pointed out that there is a typo in the OP.

Avg. Velocity for first two carriages (Va) = 40 m/ 2.0 s = 20 m/s
Avg. Velocity for first two carriages (Vb) = 40 m/ 2.4 s = 16.7 m/s (correct to 3 significant figures).

Vb should be the "Avg. Velocity for the second two carriages . . . "

 

[Kareemoya]

The train is decelerating uniformly, so it loses equal amounts of velocity in equal intervals of time.
In particular, it loses the same amount in the first second as in the second second. So after the first second it has slowed to the average of its speed at the start and its speed after two seconds.

Sorry can you explain what that has to do with time and why did the OP take the average of both times. can it maybe mean that its 2s for both carriages so 1 second per carriage

 

[kuruman]

Sorry can you explain what that has to do with time and why did the OP take the average of both times. can it maybe mean that its 2s for both carriages so 1 second per carriage

When an object travels at constant acceleration $a$ and is displaced by $\Delta x$ over time interval $\Delta t$, the relevant kinematic equations for the instantaneous velocity and the displacement are $$\begin{align}
& v= v_0+a(\Delta t) \\
& \Delta x= v_0\Delta t+\frac{1}{2}a(\Delta t)^2.
\end{align}$$ Here $v_0$ represents the velocity at the beginning of the time interval $\Delta t.$ When we divide both sides of equation (2) by $\Delta t$, we get $$\begin{align}
& \frac{\Delta x}{\Delta t}= v_0+a\left(\frac{1}{2}\Delta t\right). \end{align} $$We see that the left-hand side of equation (3) is distance divided by time a.k.a. average velocity. We recognize the right-hand side of equation (2) as the instantaneous velocity at the middle of the time interval $\Delta t.$ In other words,

The average velocity over a time interval $\Delta t$ is equal to the instantaneous velocity at the midpoint of $\Delta t$.​

When the first two cars pass the student,
$\Delta t_1=2~$s and $\Delta x_1=40~$m.
The average velocity is $40/2 = 20.0~$m/s.
The instantaneous velocity is 20.0 m/s at the midpoint of $\Delta t_1$ which is 1 s.

When the second two cars pass the student,
$\Delta t_2=2.4~$s and $\Delta x_2=40~$m.
The average velocity is $40/2.4 = 16.7~$m/s.
The instantaneous velocity is 16.7 m/s at the midpoint of $\Delta t_2$ which is 1.2 s.

Now to find the acceleration, we imagine a clock that reads $t=0$ when the front of the first car passes the student. The instantaneous velocity is $v_a=20~$m/s when the clock reads $t_a=1.0~$s. The instantaneous velocity is $v_b=16.7~$m/s when the clock reads $t_b=2.0+1.2=3.2~$s.

We can now find the acceleration from the change in instantaneous velocity. $$a=\frac{v_b-v_a}{t_b-t_a}=\frac{(16.7-20.0)~\text{m/s}}{(3.2-1.0)~\text{s}}=-1.5~\text{m/s}^2.$$ Knowing the acceleration, we can proceed to find the velocity $v_0$ of the front of the train when it passes the student.

A more straightforward approach would be to note that we have three pairs of times and positions for the front of the train relative to the student.
$x_0=0$ at $t_0=0$.
$x_1=40.0$ m at $t_1=2$ s. (Two car lengths)
$x_2=80.0$ m at $t_2=4.4$ s. (Four car lengths)

We then write the position of the front at the last two clock times $$ \begin{align} & x_1=v_0t_1+\frac{1}{2}at_1^2 \nonumber \\
& x_2=v_0t_2+\frac{1}{2}at_2^2 \nonumber \end{align}$$ and solve the system of two equations and two unknowns, $a$ and $v_0.$ The stopping distance can be found from $$x_{stop}=-\frac{v_0^2}{2a}.$$

(Edited to fix typos.)

 

[Kareemoya]

Okayyy, yeah it makes a whole lot of sense now. Couldn't thank you enough!