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October 21st, 2019, 09:06 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  How can I tell on which time both objects meet?
The problem is as follows: A kid is located at $20$ meter distance from a school bus. Once the child starts running to the bus at $6 \frac{m}{s}$, this moves from rest with a constant acceleration of $1 \frac{m}{s^{2}}$. Indicate which of the following statements from below are true (T) or false (F).I. The kid meets the bus when $t = 6\,s$. II. The kid at $t=6\,s$ is at a minimum distance of $\textrm{2 m}$ from the bus. III. The kid at $t=6\,s$ is ahead of the bus by $2\,m$. The alternatives given on my book are as follows: $\begin{array}{ll} 1.&\textrm{FFF}\\ 2.&\textrm{FTT}\\ 3.&\textrm{FTF}\\ 4.&\textrm{TFT}\\ 5.&\textrm{TTT} \end{array}$ For this problem I assumed that the kid is running at a constant speed of $6 \frac{m}{s}$: Therefore if $x(0)=0$ so he starts running from a reference point as zero: $x(t)=6t$ And for the bus it would be (as $v_{o}=0$ and $a=1$): $x(t)=20+\frac{1}{2}t^2$ Then from this the only thing which could be logical would be equating both expressions to obtain the time when both meet. But when I attempted to do so I obtained this: $6t=20+\frac{1}{2}t^2$ $12t=40+t^2$ $t^212t+40=0$ But the problem arises as the later equation doesn't yield results as real numbers. What could I be not understanding it correctly?. Now I'm assuming that the intended purpose from whoever posed this problem was that the kid is running from let's say left to right to meet the bus, and the bus is moving from the other end to meet the child. In other words from the right to the left. But would this assumption be correct?. Would it affect if it would be otherwise. I mean if the child runs to meet the bus, but the bus which is ahead by 20 meters starts moving on the same direction as the kid. Can they both meet given the latter condition?. Can somebody help me with this problem? 
October 21st, 2019, 09:44 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
No real solution for $6t = \dfrac{t^2}{2} +20$ is telling you something ... think about their respective position graphs. distance between the two is ... $d = \left(\dfrac{t^2}{2}+20\right)  6t$ How would you go about determining the minimum distance? 
October 21st, 2019, 09:47 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 
Right. There's no solution. The kid never reaches the bus. I. is False So for II. how close does the kid get? Well that's for you to determine. Set $d(t)= \dfrac{t^2}{2}  6t$ and find it's minimum. for III. just evaluate both $\dfrac{t^2}{2}$ and $6t$ to see if this is true or not. 

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