What is the speed of the tram?

Question

Every day on her way to school, Sania crosses a tram coming from the opposite direction, at a certain point on the street, at a fixed time. She left 12 minutes early one day and met the tram 3 minutes before the usual time. If she walks at 5.5 km/hr, what is the speed of the tram?

Answer 1

On that "special day", the girl walks for an extra $12 - 3 = 9$ minutes to get to the new meeting point. The tram has travelled for $3$ minutes less than usual to get to that new meeting point. The distance of the new meeting point from the usual one is the same for both. Therefore the tram is travelling $\frac 93 = 3$ times as fast as the girl, so its speed is $(5.5)(3) = 16.5 \mathrm{km/h}$.

Addendum: as I have been asked for a solution "by equations", I shall oblige. Referencing the equations you have written as comments to my answer (please do not delete or alter them),

The first equation you have written is correct: $d = (r_0 + r_1)t$.

The second equation is incorrect. It should be $d = r_0(t - \frac 1{20} + \frac 15) + r_1(t - \frac 1{20})$. The girl starts $12 \mathrm{ min}$ (one fifth of an hour) sooner, but meets the tram $3\mathrm{ min}$ (one twentieth of an hour) sooner also. The effects of these two are opposing - starting sooner gives her more of a head start and increases the travelled distance so you add that time, but meeting the tram sooner lessens the time and decreases the travelled distance, so you must subtract that time.

Equating those two (to eliminate $d$) immediately allows one to find the ratio between $r_0$ and $r_1$.

It is unclear why you have written the term $t$ twice in the "girl component" of your second equation causing it to cancel out.

Sometimes it's important to take a step back and look at the problem in a simple fashion. Equations may cloud the issue. If it can be solved without algebra, all the better.

Answer 2

I would appreciate if someone could check this over for me:

Consider $d$, the distance between her and the tram.

Also consider $r_0$ and $r_1$, the speeds of her and the tram.

We have (1) --> $d=(r_0+r_1)t$.

We also have (2) --> $d+\frac{1}{5}r_1=\frac{1}{5}r_0+(r_0+r_1)(t-\frac{1}{20})$

We know that $r_0=5.5$.

Now we subtract the equations from each other and plug in known values to get $r_1$.

Simplify the second equation a bit first.

$d=\frac{1}{5}(r_0-r_1)+(r_0+r_1)t-\frac{1}{20}({r_0+r_1})$.

Subtracting (1) from (2) gives $0=\frac{1}{5}(5.5-r_1)-\frac{1}{20}(5.5+r_1).$

Finally, we get $r_1,$ the speed of the tram, $\boxed{3.3}$ km/h.

Answer 3

Soumee,

I believe I may have an answer that is founded in equations based off your idea from before.

Set up a reference frame with left denoting the negative direction and right being positive. Place the girl's house at the origin. Using the 1-D kinematics equations for position, let the following being true:

a) $x_s = -(v_s)*t$ Where $x_s$ is Sania's position on the line and $v_s$ is the velocity of Sania.

b) $x_t = (v_t)*t - x_{0,t}$ Where $x_t$ is the position of the train on
the line and $v_t$ is the velocity of the train and $x_{0,t}$ is the train's position at the time Sania usually leaves her house.

If you let the two positions equal each other and solve for $x_{0,t}$, you get an equation analogous to your equation 1 from before. In the form of my equations this is:

c) $x_{0,t} = (v_s + v_t)(t_{M})$

Where $t_{M}$ is the usual meet up time

Next, set them equal again, but this time set the new meet up time to be 3 minutes behind $t_{M}$. The following equations should come:

d)

$$-(v_s)*(t_{M}-3 min)-(12 min * v_s) = (v_t)*(t_{M}-3 min) - x_{0,t}$$

This equality comes from the fact that they meet at the same position on the 1-D reference frame and for the fact that at t=0, Sania has already traveled for 12 minutes. Solving the equation for $v_t$ gives you the speed of the train. It should also be noted that both sides will contain equation c, so you can go ahead and cancel them on both sides of the second equality.

CHANGELOG:---------

CHANGE 1: ACCOUNTING FOR SIGN ERROR MADE (negative sign was changed to positive for $v_t$ so Sania and the train go in opposite directions)

CHANGE 2: ACCOUNTING FOR ERROR IN POSITION (equation d was reconfigured so that the train keeps its original position at time t=0 and accounts for the fact that for 12 minutes before t=0 , Sania was heading in the left direction).

CHANGE3: ADDED CLARIFICATION ON EQUATION d (elaborated on why the origin term $-(12 min * v_s)$ was put in and elaborated more on the choice for time (here t=0 is where we start the "clock" so to speak. Sania has already traveled a certain distance toward the meeting point at the time the train hits the point on the line where it usually is when Sania usually leaves).