Find the tangent line that passes through the origin

In summary, the problem is about a student trying to catch a bus that is moving with a constant acceleration of 0.170 m/s2. The student is initially 40.0 m away from the bus and running at a top speed of 5.0 m/s. The question is asking for the minimum speed the student must have to catch up with the bus. Using calculus, the student's position function is determined to be tangent to the bus's position function at the point of intersection, and solving for the velocity at that point gives the minimum speed of 0.17 m/s. An alternative method is also discussed.
  • #1
Gamerex
25
0

Homework Statement



This problem is 2.95 of University Physics, 11th edition.

Catching the Bus: A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s2.

There are subproblems a,b,c,d,e, and f; which I've all figured out except for f)

f) What is the minimum speed the student must have to just catch up with the bus?

Homework Equations



Let the subscript b mean bus, and let the subscript p mean student (p for pupil).

sp(0)=0 m
sb(0)=40 m

vp(t)=Unknown

The acceleration of the bus was given, I used calculus to find the velocity and position.
sb(t)=0.085t2+40 m
vb(t)=0.17t m/s
ab(t)=0.17 m/s2

The Attempt at a Solution



I modeled the problem by considering the bus and the student as point particles. The points in time where the student and the bus are at the same place are the intersections of the graphs of the position functions. The position function of the bus is known and given above, but the position function of the student is the integral of the student's velocity, which is constant.

The position function would be of the form sp(t)=v(t)*t+0

I managed to figure out that sp(t) must be the equation of the tangent line of sb(t) which passes through the origin.

How do I find the equation of the tangent line of sb(t)=0.085t2+40 m (And thus, the velocity) that passes through the origin?
 
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  • #2
Hmm, I didn't have to use calculus to solve this question. Set the displacements of the student and bus equal, then work from there.
 
  • #3
I don't know the displacement without knowing the velocity...
 
  • #4
I mean, take the equations of the displacement of the student and the bus, then set them equal.
 
  • #5
You mean like so?:
sp(t)=sb(t)

-> v(t)*t=0.085t2+40,
-> v(t) =(0.085t2+40)/t


This tells me what the velocity would be at my intersection point, but I still don't have the time, t.
 
  • #6
Yes, now what kind of equation do you have from there?
 
  • #7
It's a rational equation that gives me the velocity of the student.. If I set that equal to the velocity function of the bus, I can find the time where they intersect!

0.17t=(0.085t2+40)/t

t=21.7 s.. Thank you so much!
 
  • #8
Ha, I was actually going with a quadratic equation and setting √(v2 - 4ad) = 0, but your method works as well. =)
 

1. What is the equation for the tangent line that passes through the origin?

The equation for a tangent line passing through the origin is y = mx, where m is the slope of the tangent line.

2. How do you find the slope of the tangent line passing through the origin?

The slope of the tangent line passing through the origin can be found by taking the derivative of the equation of the curve at the point of interest and plugging in the x-coordinate of the origin.

3. Can you find the tangent line passing through the origin for any type of curve?

Yes, the tangent line passing through the origin can be found for any type of curve as long as the curve is differentiable at the point of interest.

4. What is the significance of finding the tangent line passing through the origin?

Finding the tangent line passing through the origin allows us to determine the rate of change of the curve at the point of interest, as well as the direction in which the curve is changing.

5. Are there any other methods to find the tangent line passing through the origin?

Yes, in certain cases, we can also use the slope-intercept form of a line (y = mx + b) to find the tangent line passing through the origin by setting the y-intercept (b) to 0.

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