0
$\begingroup$

The problem is as follows:

A car driving in a city has a constant velocity of $15\hat{i}-15\hat{j}\frac{m}{s}$ with respect to the ground in the highway. However at $t=0\,s$ the driver in the car sees a lady in highest floor of a nearby building dropping a bucket which was at rest. Find the instant (in seconds) from which the driver situated in the car will measure the speed of the bucket to be $-15\,i\frac{m}{s}$. You may consider $\vec{g}=-10\,\hat{j}\frac{m}{s^{2}}$

The alternatives given on my book are as follows:

$\begin{array}{ll} 1.&0.5\,s\\ 2.&1\,s\\ 3.&1.5\,s\\ 4.&2\,s\\ 5.&2.5\,s\\ \end{array}$

How am I supposed to relate the quantity of the velocity and the instant?.

The only equation which I can recall for the position of an object falling is:

$y(t)=y_{0}+v_{0y}t-\frac{1}{2}at^2$

But in this case I don't know how tall is the building neither of what else could I do or should I need to find what it is being asked.

I suspect that I should sum the vectors for the velocity of the car and that of the bucket falling and from then I could find the time. But I don't know which steps should I take to find such time. Can somebody help me here?.

$\endgroup$
2
  • $\begingroup$ Are the vectors aligned correctly? It seems as though the car is driving in the same direction as gravity... $\endgroup$
    – abiessu
    Nov 5 '19 at 4:17
  • $\begingroup$ @abiessu I just checked with the original source and yes they are aligned correctly. $\endgroup$ Nov 5 '19 at 4:21
0
$\begingroup$

If the car is moving with the velocity $15\hat i-15\hat j$ with respect to the ground, in the reference frame of the car, the bucket moves with velocity $-15\hat i+15\hat j$. There is no force in the $\hat i$ direction, just in the $\hat j$ direction. You can write $$v_i(t)=v_i(0)\\v_j(t)=v_j(0)-gt$$You have $v_j(t)=0$, $v_j(0)=15$ so $$0=15-10t$$or $t=1.5$ seconds.

$\endgroup$
7
  • $\begingroup$ Please help me with the conceptual basis of your answer. Why did you swap the signs of the bucket and be the "same" as that of the car?. Yes it is obvious that there is no force in the horizontal direction, but what about the vertical why the final speed in the vertical component is equal to zero?. Can you help me with this matter?. How do I get to the conclusion that $v_{j}(t)=0$?. $\endgroup$ Nov 5 '19 at 4:59
  • $\begingroup$ I left a comment which contains a question. Would you mind a clarification of my doubts please? $\endgroup$ Nov 5 '19 at 7:45
  • 1
    $\begingroup$ Answering the conceptual question: suppose you are in a fixed reference frame, and there is another reference frame, moving with velocity $\vec v_R$. If you have a object in the moving reference frame with velocity $\vec v_1$ in that frame, you are going to see it moving with velocity $\vec v_0=\vec v_1+\vec v_R$. Think of throwing a ball in a train. You know that in your reference frame the bucket is at rest, so $\vec v_0=0$. Then $\vec v_1=-\vec v_R$. As for your 2nd question, the problem asked about the time when the measured velocity is only in the $\hat i$ direction, so no $j$ omponent. $\endgroup$
    – Andrei
    Nov 5 '19 at 12:53
  • $\begingroup$ Can you offer a second example so I could understand it better?. I mean about reference frames. In your answer you mentioned about a ball and then a bucket. Are you referring to a particular situation of your example or what it is in the problem?. $\endgroup$ Nov 7 '19 at 8:25
  • $\begingroup$ For the second question, I forgot to ask that you mention there is no force in the $i$ component is it because there isn't any gravity acting there or what?. I'm assuming that you set$j=0$ because for the driver in the car he will not see the object falling? or from his point of view he will only notice the moving in the $i$ frame?. This part is where I'm confused at. Can you help me with that please?. $\endgroup$ Nov 7 '19 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.