- #1

Aichuk

- 29

- 1

**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

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.

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

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 :)

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 :)