My Math Forum  

Go Back   My Math Forum > Science Forums > Physics

Physics Physics Forum


Thanks Tree2Thanks
  • 1 Post By skeeter
  • 1 Post By DarnItJimImAnEngineer
Reply
 
LinkBack Thread Tools Display Modes
October 21st, 2019, 08:39 PM   #1
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Question How do I find the instant when two moving objects are together using a graph?

The problem is as follows:
The graph from below describes the motion of an electric car and a diesel truck. It is known that they pass through the same point at $t=0$. Find the instant when they are together again.


The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{12 s}\\
2.&\textrm{15 s}\\
3.&\textrm{18 s}\\
4.&\textrm{16 s}\\
5.&\textrm{20 s}
\end{array}$

For this problem the only thing that I could come up with was to identify the two equations for speed for both the truck and the electric car which are as follows. I'm using the labels $\textrm{t=truck and c=electric car}$

$v_{t}(t)=10$

$v_{c}(t)=2t$

from equating both I obtainted the time when they do have the same speed. (For brevity purposes I'm omitting the units but they are consistent)

$2t=10$

$t=5$

But from then on I don't know what to do with that information or how do I join it with other equation to get to the time when they are together.

Can somebody help me with this?.
Chemist116 is offline  
 
October 21st, 2019, 09:26 PM   #2
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,101
Thanks: 1677

Area under the velocity graph is displacement ...

truck displacement = car displacement

$\displaystyle \int_0^t 10 \, dt = \int_0^6 2t \, dt + \int_6^t 12 \, dt$

$10t = 36 + (12t-72)$

$10t = 12t -36 \implies t=18$
skeeter is offline  
October 22nd, 2019, 06:37 AM   #3
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,101
Thanks: 1677

w/o calculus

for $0 \le t \le 6$ ...

$v_c = 2t \implies x_c(6) = \dfrac{1}{2}(2) \cdot 6^2 = 36$

$v_T = 10 \implies x_T(6) = 10(6) = 60$

truck is 24 meters ahead at time $t=6$

for $t > 6$

$24 + 10t = 12t \implies t = 12$

car passes truck at time $t= 6 + 12$ sec.
skeeter is offline  
October 28th, 2019, 09:14 PM   #4
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Question

Quote:
Originally Posted by skeeter View Post
w/o calculus

for $0 \le t \le 6$ ...

$v_c = 2t \implies x_c(6) = \dfrac{1}{2}(2) \cdot 6^2 = 36$

$v_T = 10 \implies x_T(6) = 10(6) = 60$

truck is 24 meters ahead at time $t=6$

for $t > 6$

$24 + 10t = 12t \implies t = 12$

car passes truck at time $t= 6 + 12$ sec.
The calculus approach that you used is pretty straightforward and I'm familiar with it at some extend but the approach you used here I'm still stuck at where does it come this part?.

$\dfrac{1}{2}(2) \cdot 6^2 = 36$

Are you implying to use the known areas by the area of the right triangle but that wouldn't be?

$A=\frac{bh}{2}$

I'd like you could explain me this part please.
Chemist116 is offline  
October 29th, 2019, 04:50 AM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 21,124
Thanks: 2332

The sketch isn't to scale.

In the first 6 s, the car's average velocity is 6 m/s,
so the truck gets ahead by 6(10 - 6) m = 24 m.
After 6 seconds, the car's velocity relative to the truck is a constant 2m/s,
so the car catches up with the truck after a further 12 seconds.

Last edited by skipjack; October 29th, 2019 at 05:14 AM.
skipjack is offline  
October 29th, 2019, 08:00 AM   #6
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 3,101
Thanks: 1677

Velocity graphs to scale.

Compare the area of the red trapezoid to the blue rectangle
Attached Images
File Type: jpg Motion3.jpg (95.7 KB, 1 views)
File Type: jpg Motion4.jpg (92.7 KB, 0 views)
Thanks from Chemist116
skeeter is offline  
November 1st, 2019, 02:41 PM   #7
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Question

Quote:
Originally Posted by skipjack View Post
The sketch isn't to scale.

In the first 6 s, the car's average velocity is 6 m/s,
so the truck gets ahead by 6(10 - 6) m = 24 m.
After 6 seconds, the car's velocity relative to the truck is a constant 2m/s,
so the car catches up with the truck after a further 12 seconds.
Why the average velocity of the car is 6 m/s?.

The term average is where I'm stuck at.
The second statement I presume it comes from using $6t$.
The third part about relative velocity of the car where you mention that is constant I'm lost. What's the meaning of relative velocity and how did you obtained?.

Sorry but these look conceptual questions and I'd like some guidance because I think it is the answer of how to understand what you mentioned.
Chemist116 is offline  
November 1st, 2019, 02:56 PM   #8
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Question

Quote:
Originally Posted by skeeter View Post
Velocity graphs to scale.

Compare the area of the red trapezoid to the blue rectangle
In the approach where you didn't used calculus, I thought that you used areas but instead you used the position equation for both objects to find the separation between them.

The part where you used this equation:

$10t+24=12t$

Is only possible when the two objects are moving with no acceleration involved, in other words uniform motion, hence this is why you used the interval $t>6$.

But again, I think that for that equation from above, areas were not used. Am I right?.
Chemist116 is offline  
November 1st, 2019, 03:00 PM   #9
Senior Member
 
Joined: Jun 2019
From: USA

Posts: 386
Thanks: 211

The average value of any f(t) is $\displaystyle \bar{f} = \frac{1}{\Delta t} \int_{t_0}^{t_0 + \Delta t} f dt$.

An integral is the area under the curve. In the first 6 seconds, the average velocity is the area under the triangle divided by the width of the triangle. This is the "average height."

For a linear relation like this, it's easy. The average velocity will be $\displaystyle \frac{0~m/s+12~m/s}{2} = 6~m/s$.
Thanks from Chemist116
DarnItJimImAnEngineer is offline  
November 1st, 2019, 05:25 PM   #10
Senior Member
 
Chemist116's Avatar
 
Joined: Jun 2017
From: Lima, Peru

Posts: 188
Thanks: 5

Math Focus: Calculus
Quote:
Originally Posted by DarnItJimImAnEngineer View Post
The average value of any f(t) is $\displaystyle \bar{f} = \frac{1}{\Delta t} \int_{t_0}^{t_0 + \Delta t} f dt$.

An integral is the area under the curve. In the first 6 seconds, the average velocity is the area under the triangle divided by the width of the triangle. This is the "average height."

For a linear relation like this, it's easy. The average velocity will be $\displaystyle \frac{0~m/s+12~m/s}{2} = 6~m/s$.
That was what I needed but I am still stuck at exactly how skipjack arrives to the conclusion from the statements given that the car catches the truck at $t=12$. Is it because the two speeds are separated by $2$ thus will it take that twice $6\,s$ for the car to catch the truck?
Chemist116 is offline  
Reply

  My Math Forum > Science Forums > Physics

Tags
find, graph, instant, moving, objects



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How do I find the displacement of a particle at a given instant? Chemist116 Physics 3 October 22nd, 2019 05:36 AM
Moving Connected Graph router Topology 0 December 9th, 2014 06:16 PM
Distance between Moving Objects / graphs of equations qayan890 Algebra 1 February 15th, 2014 12:55 AM
A moving point along a graph mathkid Calculus 12 October 11th, 2012 08:17 PM
Plotting moving points in 2d graph FKeeL Algebra 0 September 10th, 2009 08:23 AM





Copyright © 2019 My Math Forum. All rights reserved.