Can Savannah Catch the Accelerating School Bus?

AI Thread Summary
Savannah is trying to catch a school bus that starts accelerating from a stop at 1.10 m/s², while she runs at a constant speed of 6.40 m/s from a distance of 30.0 m. The discussion revolves around using kinematic equations to determine if she can catch the bus and, if not, how close she gets. After setting up the equations for both Savannah and the bus, it was concluded that Savannah does not catch the bus, as the quadratic equation derived from their distances has no real roots. The key takeaway is to find the point when the bus's velocity exceeds Savannah's, indicating she will fall further behind.
sbayla31
Messages
9
Reaction score
0

Homework Statement



My teacher gave our class a set of questions that are supposed to be challenging. This question has particularly stumped me... I can make a diagram, but I really don't know what to do next:

Savannah has just stepped out her front door, only to find that the school bus, located at the end of her driveway 30.0 m from her door, has just begun to accelerate away from her at 1.10 m/s2. Savannah immediately begins to chase the school bus at her top speed of 6.40 m/s. Will she catch the bus? Yes- where/when does she catch up to it? No- how close will she get?

Homework Equations



These are just guesses.

c2=a2+b2

<br /> v_{ave} = \Delta x / \Delta t<br />

The Attempt at a Solution



Savannah:
v1=6.40 m/s
a=0
dx= ?
dy= 30.0m
t= ?

Am I on the right track? :confused:

I know you guys want me to try to solve the problem, but I've gotten a headache thinking about it :frown:
 
Physics news on Phys.org
Hello sbayla31,
Welcome to Physics Forums!
sbayla31 said:

Homework Statement



My teacher gave our class a set of questions that are supposed to be challenging. This question has particularly stumped me... I can make a diagram, but I really don't know what to do next:

Savannah has just stepped out her front door, only to find that the school bus, located at the end of her driveway 30.0 m from her door, has just begun to accelerate away from her at 1.10 m/s2. Savannah immediately begins to chase the school bus at her top speed of 6.40 m/s. Will she catch the bus? Yes- where/when does she catch up to it? No- how close will she get?

Homework Equations



These are just guesses.

c2=a2+b2

<br /> v_{ave} = \Delta x / \Delta t<br />

The Attempt at a Solution



Savannah:
v1=6.40 m/s
a=0
dx= ?
dy= 30.0m
t= ?

Am I on the right track? :confused:

I know you guys want me to try to solve the problem, but I've gotten a headache thinking about it :frown:
There is a set of equations given in your coursework called the "Kinematics equations for uniform acceleration." There should be at least 4 of them. I strongly suggest memorizing these equations. (There are surprisingly very few equations that need to be memorized when learning physics. But these are some of the few that are really good to memorize.) Having these equations handy will work as good headache medicine in the future. :smile:

I'll present a couple of them here, but you should memorize them all.

\vec s = \vec s_0 + \vec v_0 t +\frac{1}{2} \vec a t^2

\vec a = \frac{\vec v_f - \vec v_i}{t}

Your coursework/textbook might use different notation. I suggest using the notation used by your coursework.

Create equations for both the bus's distance from the house and Savannah's distance from the house (each as a function of time, t). Also create an equation for the velocity of the bus as a function of time, t.

When does the bus's velocity equal Savannah's velocity? At this time, how far is Savannah from her house? How far is the bus from the house?

(At that time, if Savannah is farther from her house from the bus is from her house, it really means that she caught up with the bus at some time earlier. Find the time when the distance of Savannah from her house is equal to the distance of the bus from the house.)
 
Thank you for your help :)
I discussed the problem with some friends and now I have this:

Savannah

d=v1t+1/2at2
d=6.4t+0 (no acceleration)
d=6.4t

Bus

d=v1t+1/2at2+30
d=(1/2)(1.1)t2+30
d=0.55t2+30

Savannah = Bus

6.4t=0.55t2+30
0.55t2-6.4t+30=0

and I did the quadratic equation and there were no real roots, so I know that she does not catch up to the bus.

So how do I know how close to the bus she gets?
 
sbayla31 said:
So how do I know how close to the bus she gets?

We know that the bus started out from a stop, and Savannah started out at a constant 6.4 [m/s]. As long as the bus's velocity is less than Savannah's, Savannah is able to catch up a little bit. But as soon as the bus's velocity becomes greater than Savannah's, she'll keep falling further behind.

So when is the bus's velocity equal to Savannah's? And at what net distance is she to the bus?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top