What is the maximum upward acceleration of the ball?

In summary, you are having trouble with 3 of your 10 problems. You do calculations and everything, but you always seem to come up with the wrong answer. If anyone can help you out, that would be great.
  • #1
Effitol840
16
0
Alright I'm having trouble with 3 of my 10 problems I have for homework. I do calculations and everything but I always seem to come up with the wrong answer. If anyone can help me out that wold be great. The problems are as follows:

1. A certain cable car is San Francisco can stop in 10 s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d m in front of the car and slams on the brakes instantly. The car reaches the dog 8.6 s later, and the dog jumps off the track just in time. If the car travels 4.5 m beyond the position of the dog before coming to a stop, how far was the car from the dog?

2. Suppose a small child rolls off a bed that is 0.44 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.9 mm. If the floor is carpeted, this stopping distance is increased to about 1.0 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume that the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.

3. A hard rubber ball, released at chest height (1.50 m), falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Before this dent in the ball pops out, suppose that its maximum depth is 1.10 centimeter. What is the maximum upward acceleration of the ball?

The following equations are supposed to be helpful with all of these problems:

[tex]v[/tex]=velocity [tex]v_{0}[/tex]=initial velocity [tex]\Delta x[/tex]=change in distance [tex]a[/tex]=accelertion
[tex]t[/tex]=time



[tex]vt=\Delta x[/tex]
[tex]v=v_{0}+at[/tex]
[tex]v^2=v_{0}^2+2a\Delta x[/tex]
[tex]\Delta x=v_{0}t+\frac{1}{2}at^2[/tex]

If anyone can help me with how to get started with these the correct way that would be great. Thanks.
 
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  • #2
You're supposed to tell us what you've already tried,
and how you're thinking about each problem
(see sticky link at very top of the college HW section).

1. you should presume constant acceleration, so
at the dog the trolley has speed 1.4/10 of v_max.
The average speed after the dog was ? of v_max ;
you know how far it travels in 1.4 sec at this speed.
The average speed before the dog was ?? v_max ;

in (2), you know 2ax before hitting the floor;
so you figure out v^2 before the floor; you're given x's!

in (3), again you're given x that achieves v_f = 0.
just plug it in!
Don't worry much about it - few students understand
the v^2 formula at first ... this is okay I suppose,
since it doesn't really fit in with Newton's view of motion.
After your first exam you'll see it in a new guise,
(KE) where it will make perfect sense, fitting in
with Leibnitz' and Hamilton's approach to dynamics.
 
  • #3
Sorry about not including information with the problems. I've tried way too many things to write them down. I understand the concept of using and re-using and contorting the equations as to get the answer that I need, I guess I really don't know where I go wrong.

1) With the first problem I figured that the acceleration was constant and that since it is constant it would have the same slope all the way through. I got -1.15 [tex]m/s^2[/tex] from using the equation [tex]a=\frac{v^2-v_0^2}{2\Delta x}[/tex] which I got from one of my original equations. Thats where I get lost. I think its because that's not really the accelleration. I used v from finding it between the end of the line between 8.6s and 10s. So wrong or right I'm still having trouble with the rest of the problem.

2 & 3) As for these 2 problems I have no clue where to start, partially because I really can't understand what the question is asking.

So I guess my real question is how to start 2 and 3 and also what to do to finish 1.

-Effitol840
 
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Related to What is the maximum upward acceleration of the ball?

1. What is the definition of maximum upward acceleration?

The maximum upward acceleration is the highest rate at which the velocity of an object increases in an upward direction.

2. How is the maximum upward acceleration calculated?

The maximum upward acceleration can be calculated using the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What factors affect the maximum upward acceleration of a ball?

The maximum upward acceleration of a ball can be affected by factors such as the ball's mass, the force applied to the ball, and air resistance.

4. Can the maximum upward acceleration of a ball change?

Yes, the maximum upward acceleration of a ball can change depending on the forces acting upon it. For example, if a stronger force is applied to the ball, its maximum upward acceleration will increase.

5. Why is it important to understand the maximum upward acceleration of a ball?

Understanding the maximum upward acceleration of a ball is important in many scientific fields, such as physics and sports science. It allows us to predict the trajectory and behavior of the ball, and also helps in designing equipment and structures that can withstand the forces involved.

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