Can Savannah Catch the Accelerating School Bus?

In summary, Savannah chases the school bus at her top speed of 6.40 m/s. She catches up to the bus at some time before it reaches the end of the driveway, but she never reaches it.
  • #1
sbayla31
9
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

[tex]
v_{ave} = \Delta x / \Delta t
[/tex]

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:
 
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  • #2
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

[tex]
v_{ave} = \Delta x / \Delta t
[/tex]

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.

[tex] \vec s = \vec s_0 + \vec v_0 t +\frac{1}{2} \vec a t^2 [/tex]

[tex] \vec a = \frac{\vec v_f - \vec v_i}{t} [/tex]

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.)
 
  • #3
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?
 
  • #4
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?
 
  • #5


As a scientist, it is important to approach problems with a clear and focused mindset. It is understandable that this question may have stumped you, but don't let it discourage you. First, it is important to understand what is given in the problem and what is being asked. In this case, we have the initial distance between Savannah and the school bus, the acceleration of the bus, and Savannah's top speed. We are being asked if she will catch the bus and if so, where/when she will catch it.

To solve this problem, we can use the equation for position, velocity, and acceleration in kinematics: x = x0 + v0t + 1/2at^2. In this case, x0 is the initial position (30.0 m), v0 is the initial velocity (0 m/s), a is the acceleration of the bus (1.10 m/s^2), and t is the time. We can also use the fact that Savannah's velocity is constant (6.40 m/s) to find the time it takes her to catch the bus.

To find the time, we can set the equations for Savannah's position and the bus's position equal to each other and solve for t. This will give us the time at which Savannah catches the bus. We can then plug this time into the equation for Savannah's position to find the distance she has traveled when she catches the bus. If the distance is less than 30.0 m, then she will catch the bus. If the distance is greater than 30.0 m, she will not catch the bus.

I understand that this may seem overwhelming, but don't give up! Take a break and come back to the problem with a fresh mind. You can also ask your teacher or classmates for help or clarification. Remember, as a scientist, it is important to persist in solving problems and to ask for help when needed. Good luck!
 

Related to Can Savannah Catch the Accelerating School Bus?

1. What is a kinematics diagram problem?

A kinematics diagram problem is a type of physics problem that involves analyzing the motion of objects using a diagram. It is commonly used in mechanics and engineering to understand the movement of objects in a system.

2. How do I draw a kinematics diagram?

To draw a kinematics diagram, you will need to identify the objects in the system, the forces acting on them, and the direction of motion. Draw each object as a point or a simple shape, and label the forces and their directions. Make sure to include a scale and coordinate axes for accuracy.

3. What are the key equations used in kinematics diagram problems?

The key equations used in kinematics diagram problems are the equations of motion, which include displacement, velocity, acceleration, and time. These equations are used to describe the motion of objects in a system and can be derived from the fundamental laws of motion.

4. How do I solve a kinematics diagram problem?

To solve a kinematics diagram problem, you will need to follow a systematic approach. Start by drawing the diagram and labeling all known values. Then, use the equations of motion to solve for the unknown values. Finally, check your answer for accuracy and make any necessary adjustments.

5. What are some common mistakes to avoid when solving kinematics diagram problems?

Some common mistakes to avoid when solving kinematics diagram problems include using incorrect equations, not considering all forces acting on an object, and not using proper units throughout the problem. It is also important to double-check your calculations and make sure they are consistent with the given information.

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