Solving a Puzzling Physics Problem: ##15 \pm 3\sqrt{15}##

In summary, the problem involves two cars starting at rest and traveling in the same direction. The first car has an acceleration of 3 m/s^2 and the second car has an acceleration of 5 m/s^2. The question asks for the time it takes for the second car to overtake the first. The solution involves finding the distance traveled by each car and setting them equal to each other. The final answer is ~26.6 seconds.
  • #1
ryanpruette
3
1
Homework Statement
Two cars start at rest at the same point and travel in the same direction. The first car leaves at ##t_1 = 0## with an acceleration of ##a_1 = 3 m/s^2##. The second car leaves at ##t_2 = 6## with an acceleration of ##a_2 = 5 m/s^2##. At what time does the second car overtake the first?
Relevant Equations
$$v_x = v_{0x} + \int_0^t a_x \, dt$$
$$x = x_0 + \int_0^t v_x \, dt$$
So this isn't actually a homework question, it's a question that was on the test I just took and I cannot figure out what I'm doing wrong. I've tried solving it three or four different ways, double checking everything, and I keep getting ##15 + 3\sqrt {15}## or ~26.6, which was not one of the answer choices.

Starting by finding ##x_1(t),## I know initial position and velocity are both zero so I can find ##x_1(t)## by integrating ##a_1 = 3 m/s^2## twice. $$v_1(t) = \int_0^t 3 \, dt = 3t$$
$$x_1(t) = \int_0^t 3t \, dt = \frac 3 2 t^2$$

To find ##x_2(t),## I did the same thing but I had to find initial position and velocity as well.
$$v_2(t) = v_0 + \int_0^t 5 \, dt$$
$$v_2(t) = v_0 + 5t$$
I know ##v_2(6) = 0,## so
$$0 = v_0 + 5(6)$$
$$v_0 = -30$$
Therefore, $$v_2(t) = 5t - 30$$
Then, I did the same thing again to find ##x_2(t)##
$$x_2(t) = x_0 + \int_0^t 5t - 30\, dt$$
$$x_2(t) = x_0 + \frac 5 2 t^2 - 30 t$$
$$ 0 = x_0 + \frac 5 2 (6)^2 - 30 (6)$$
$$0 = x_0 + 90 - 180$$
$$x_0 = 90$$
$$x_2(t) = \frac 5 2 t^2 - 30t + 90$$

Finally, to find the final answer, I set them equal to each other and solved for t
$$\frac 3 2 t^2 = \frac 5 2 t^2 - 30t + 90$$
$$0 = t^2 - 30t + 90$$
$$t = \frac {30 \pm \sqrt{30^2 -4(90)}} {2},$$
$$t = 15 \pm 3 \sqrt 15$$
$$t = 3.38, t = 26.6$$
And I threw out ##t = 3.38## because it's less than 6.
 
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  • #2
ryanpruette said:
v2(t)=v0+∫0t5dt
v2(t)=v0+5t
We know x_2=0, v_2=0 for 0<t<6 . The above does not satisfy it.
 
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  • #3
ryanpruette said:
$$v_0 = -30$$
At what point in the proceedings is car 2 reversing??
 
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  • #4
ryanpruette said:
The first car leaves at ##t_1 = 0## with an acceleration of ##a_1 = 3 m/s^2##. The second car leaves at ##t_2 = 6## with an acceleration of ##a_2 = 5 m/s^2##.
Hello.
For 0<t<6 we know that first car is moving with constant acceleration but second car is at rest.
So for the first car find it's position and velocity at t=6s then you know velocity and position of both cars at t=6s.
For first car you calculate using x=0.5a(t^2)+v0t+x0 and v=at+v0 and for second car which was at rest you know that at t=6s it's position is still x=0 and it's velocity is 0.
I hope it helps ...
 
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  • #5
ryanpruette said:
Homework Statement:: Two cars start at rest at the same point and travel in the same direction. The first car leaves at ##t_1 = 0## with an acceleration of ##a_1 = 3 m/s^2##. The second car leaves at ##t_2 = 6## with an acceleration of ##a_2 = 5 m/s^2##. At what time does the second car overtake the first?
Relevant Equations:: $$v_x = v_{0x} + \int_0^t a_x \, dt$$
$$x = x_0 + \int_0^t v_x \, dt$$
Hello @ryanpruette ,

:welcome:

Is that the exact wording of the problem? Might the question have asked, "How long does it take the second car to overtake the first?" or something to that effect.

I agree with your answer. (Your method for car 2 is a bit strange.)

Starting from rest, car 1 travels the same distance in ##15 + 3\sqrt {15}## seconds as car 2 travels in ##9 + 3\sqrt {15}## seconds.
 
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  • #6
ryanpruette said:
, and I keep getting ##15 + 3\sqrt {15}## or ~26.6, which was not one of the answer choices.
I too agree with your answer (except mine has units!). What is the ‘official’ answer?

A few general thoughts which might be useful…

You have overcomplicated the calculation for car-2.

Suppose the cars are level at time t= T. Car-1 has accelerated for a time of duration T. Car-2 has accelerated for a time of duration T-6.

The distance covered by car-1 while accelerating during T can be found by integrating from 0 to T as you have done.

In the same way, the distance covered by car-2 while accelerating during T-6 can be found by integrating from 0 to T-6. Then there’s no need to find ##v_0## and ##x_0## for car-2. In this particular problem it's far more convenient to take car-2’s time-variable as being zero when car-2 starts moving.

Also, for constant acceleration problems such as this, we don’t usually use calculus. We use the standard equations of constant acceleration. For example, look-up ‘suvat’ equations. With this approach we can immediately write down ##\frac 12 * 3 * T^2 = \frac 12 * 5 * (T-6)^2## and all that remains is to solve the quadratic equation.

[Minor edits]
 
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  • #7
SammyS said:
Hello @ryanpruette ,

:welcome:

Is that the exact wording of the problem? Might the question have asked, "How long does it take the second car to overtake the first?" or something to that effect.

I agree with you answer. (Your method for car 2 is a bit strange.)

Starting from rest, car 1 travels the same distance in ##15 + 3\sqrt {15}## seconds as car 2 travels in ##9 + 3\sqrt {15}## seconds.
Ohhhh, that must be the issue. I must have been so focused on finding the correct value of t, I didn't think to make sure I was interpreting the question correctly. Iirc 21 s was an answer choice which would've been rounded from ##9 + 3\sqrt {15}##. Thanks for the clarity, I have sadly been thinking about this question since Wednesday lol
 
  • #8
anuttarasammyak said:
We know x_2=0, v_2=0 for 0<t<6 . The above does not satisfy it.
Yeah, ig the correct way to write it would've been$$
x_2(t) =
\begin{cases}
0, & x \lt 6 \\
\frac 5 2 t^2 - 30 t + 90, & x \geq 6
\end{cases}
$$
but you still get the same answer, no?
 
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  • #9
ryanpruette said:
Yeah, the correct way to write it would've been$$
x_2(t) =
\begin{cases}
0, & x \lt 6 \\
\frac 5 2 t^2 - 30 t + 90, & x \geq 6
\end{cases}
$$
but you still get the same answer, no?
Yes, you get the same result for ##t\ge 6## for car 2, so you get the same answer.
 
  • #10
ryanpruette said:
Ohhhh, that must be the issue. I must have been so focused on finding the correct value of t, I didn't think to make sure I was interpreting the question correctly. Iirc 21 s was an answer choice which would've been rounded from ##9 + 3\sqrt {15}##. Thanks for the clarity, I have sadly been thinking about this question since Wednesday lol
Please, see attached diagram.
 

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  • #11
"I cannot figure out what I am doing wrong"

"I know v_2(6)=0, so
0=v_0+5(6)
v_0=−30 ".

Here you are assuming
(1) acceleration to be 5 during the first 6 sec. During the first 6 sec acceleration actually is 0.
(2) You are assuming the initial velocity of second car to be unknown but that is known. It is 0.
For the first 6 sec the second car is at rest so its velocity and acceleration both are 0.
 
  • #12
physicalbug said:
"I cannot figure out what I am doing wrong"

"I know v_2(6)=0, so
0=v_0+5(6)
v_0=−30 ".

Here you are assuming
(1) acceleration to be 5 during the first 6 sec. During the first 6 sec acceleration actually is 0.
(2) You are assuming the initial velocity of second car to be unknown but that is known. It is 0.
For the first 6 sec the second car is at rest so its velocity and acceleration both are 0.
@ryanpruette confused everybody by not explaining his method. He set up the equation of position for the second car so that it would be correct for t>6$$x_2(t) = \frac 5 2 t^2 - 30t + 90$$.
That works.
 
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  • #13
haruspex said:
@ryanpruette confused everybody by not explaining his method. He set up the equation of position for the second car so that it would be correct for t>6$$x_2(t) = \frac 5 2 t^2 - 30t + 90$$.
That works.
MatinSAR has given the simplest way to solve this. Taking initial velocity of the second car to be -30 m/s is wrong. That value is 0.
 
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  • #14
physicalbug said:
MatinSAR has given the simplest way to solve this. Taking initial velocity of the second car to be -30 m/s is wrong. That value is 0.
It's not quite wrong, but it is a solution out of left-field and it does need some explanation.

It's a bit like the "method of images" in EM, where you construct an alternative problem that has the same solution as the original problem. Then you solve the alternative problem.

If the second car had an initial position of ##+90m## and an initial velocity of ##-30 m/s## and a constant acceleration of ##+5m/s^2##, then it would have the same motion from ##t = 6s## as in the real problem(!) So, if we ignore any solutions with ##t < 6s##, we must get the same solution as in the real problem.
 
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  • #15
physicalbug said:
MatinSAR has given the simplest way to solve this. Taking initial velocity of the second car to be -30 m/s is wrong. That value is 0.
Hello.
This is how I have tried to solve the problem. With knowing that :
##x=x_{0}+v_{0}(t-t_{0})+\frac 1 2 a(t-t_{0})^2##​
If you have ##t_{0}=0## for arbitrary ##x_{0}## then you have :
##x=x_{0}+v_{0}t+\frac 1 2 at^2##
Way 1:
For car 1 we have : ##t_{0}=0 , x_{0}=0 , v_{0}=0 , a=3## so :
##x_{1}=\frac 3 2 t^2##​
For car 2 we have : ##t_{0}=6 , x_{0}=0 , v_{0}=0 , a=5## so :
##x_{2}=\frac 5 2 (t-6)^2## for ##t>6##
At what time does the second car overtake the first? We need to solve ##x_{1}=x_{2}=A## :
##A## is the position where the second car reaches the first car.
##\frac 3 2 t^2=\frac 5 2 (t-6)^2##​
Solve for ##t## : (We accept only ##t>6## and it is obvious why ...)
##t=15+3\sqrt15## True!
##t=15-3\sqrt15## False!

##t=15+3\sqrt15## is the time that car 1 reaches ##x=A##.
##t=15+3\sqrt15 -6 = 9+3\sqrt15## is the time that car 2 reaches ##x=A##.

Way 2 :
For car 2 we have : ##t_{0}=0 , x_{0}=0 , v_{0}=0 , a=5## so :
##x_{2}=\frac 5 2 t^2##​
For car 1 we have : ##t_{0}=-6 , x_{0}=0 , v_{0}=0 , a=3## so :
##x=\frac 3 2 (t+6)^2## for ##t>-6##​

Continue like before, you get ##t=9+3\sqrt15## which is the time that car 2 reaches car 1.
 
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  • #16
Here's a Desmos approach:

Note the use of piecewise-defined functions (as suggested earlier by @ryanpruette).
Note the use of regression (via the ##\tilde{\phantom{w}}## symbol) and the suggestion to probe the range ##T\geq 6##.

https://www.desmos.com/calculator/itf6nyxcub

1676308982763.png


You can modify the functions to not-zero-out the parabolas before the vertex,
and possibly give suggested ranges for T in the regression.

You can also view the velocity functions.
You might have to zoom in (on the vertical axis) to see them.

Of course, while the goal is to solve the problem analytically,
seeing the graphs might help give some intuition about the physical and unphysical solutions.
 
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  • #17
physicalbug said:
MatinSAR has given the simplest way to solve this. Taking initial velocity of the second car to be -30 m/s is wrong. That value is 0.
It depends what you mean by initial here. @ryanpruette's method takes the initial state as that at t=6. Finding a value for velocity at t=0 was merely a step to finding the general equation for t>6.
It's as simple as any, when you understand it.
 
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  • #18
ryanpruette said:
Yeah, ig the correct way to write it would've beenx2(t)={0,x<652t2−30t+90,x≥6
but you still get the same answer, no?
The problem states
[tex]a(t)=5\theta(t-6),\ v(0)=0,\ x(0)=0[/tex]
where step function ##\theta(y)##=0 for y<0, 1 for y>0.
Integrating it with keeping continuity at t=6,
[tex]v(t)=5(t-6)\ \theta(t-6)[/tex]
[tex]x(t)=\frac{5}{2}(t-6)^2\ \theta(t-6)[/tex]

"If there exists a car #3 which keep acceleration 5 all the time and meet car #2 when t=6 at x=0 with v=0, then their motion at t>6 coincide. So let us investigate motion of car #3 instead, because not-constant acceleration is difficult to handle". This is my interpretaion of your motif and method. For car #3
[tex]v(t)=5(t-6)[/tex]
[tex]x(t)=\frac{5}{2}(t-6)^2[/tex]
for all the time.
 
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Related to Solving a Puzzling Physics Problem: ##15 \pm 3\sqrt{15}##

What does the expression ##15 \pm 3\sqrt{15}## represent in a physics context?

The expression ##15 \pm 3\sqrt{15}## typically represents a solution to a quadratic equation or a value derived from a physical problem involving quadratic relationships, such as kinematic equations, energy levels in quantum mechanics, or other scenarios where quadratic forms arise.

How do you derive the expression ##15 \pm 3\sqrt{15}## from a quadratic equation?

To derive the expression ##15 \pm 3\sqrt{15}## from a quadratic equation, you would start with a standard form quadratic equation ax^2 + bx + c = 0 and use the quadratic formula x = (-b ± √(b² - 4ac)) / 2a. By identifying appropriate values for a, b, and c, you can solve for x to obtain the expression.

What are the possible values of the expression ##15 \pm 3\sqrt{15}##?

The expression ##15 \pm 3\sqrt{15}## yields two possible values: 15 + 3√15 and 15 - 3√15. These values are the solutions to the quadratic equation or the specific physical quantities being calculated.

In what types of physics problems might the expression ##15 \pm 3\sqrt{15}## appear?

The expression ##15 \pm 3\sqrt{15}## might appear in physics problems involving quadratic relationships. Examples include projectile motion (where the displacement equation is quadratic), energy quantization in quantum mechanics, and solving for characteristic polynomial roots in systems of differential equations.

Can the expression ##15 \pm 3\sqrt{15}## be simplified further?

No, the expression ##15 \pm 3\sqrt{15}## is already in its simplest form. It represents two distinct numerical values, and further simplification is not possible without approximating the square root of 15.

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