# Need Help Solving Physics Practice Problems

• mcr
In summary: I have d=14.67(ft/sec)t+4.2 and for the other runner I have d=1/2at^2. I'm not sure on these. That's all I have.Not quite. For t<2 what does the graph of velocity vs time look like for the second runner? What is the area under the graph for that part? What does the graph look like for t>=2? The area under the graph for t<=2 is going to be smaller than the area under the graph for t>=2 because the fast runner has been moving for a shorter time. The area
mcr
Hi,
I'm new at this and I need some help with the following practice problems:

1) At what angle relative to the horizontal must an object be
launched if its minimum velocity in flight is 28% of its launch
velocity? The answer is to be in degrees. I have the answer to be 73.7.

2) During a footrace, a fast runner runs at a 6 min/mi pace, and a
slow runner runs at a 7 min/mi pace. The fast runner passes the slow
runner. After a 2 sec delay, the slow runner initiates an acceleration,
catching up with the fast runner in 968 ft. What is this acceleration?
The answer is to be in ft/sec^2 and the answer is .0798.

I'm confused on both and don't know how to get started.

welcome to pf!

hi mcr! welcome to pf!
mcr said:
1) At what angle relative to the horizontal must an object be
launched if its minimum velocity in flight is 28% of its launch
velocity? The answer is to be in degrees. I have the answer to be 73.7.

yup!
2) During a footrace, a fast runner runs at a 6 min/mi pace, and a
slow runner runs at a 7 min/mi pace. The fast runner passes the slow
runner. After a 2 sec delay, the slow runner initiates an acceleration,
catching up with the fast runner in 968 ft. What is this acceleration?
The answer is to be in ft/sec^2 and the answer is .0798.

I'm confused on both and don't know how to get started.

write two equations, one for each runner, both starting at the time that they are level

(so the second one will have constant speed for t ≤ 2, and https://www.physicsforums.com/library.php?do=view_item&itemid=204" for t > 2 )

Last edited by a moderator:
mcr said:
Hi,
I'm new at this and I need some help with the following practice problems:

1) At what angle relative to the horizontal must an object be
launched if its minimum velocity in flight is 28% of its launch
velocity? The answer is to be in degrees. I have the answer to be 73.7.
What is the condition for minimum total velocity? (hint: does the horizontal velocity change during flight - ignoring air resistance -?)

2) During a footrace, a fast runner runs at a 6 min/mi pace, and a
slow runner runs at a 7 min/mi pace. The fast runner passes the slow
runner. After a 2 sec delay, the slow runner initiates an acceleration,
catching up with the fast runner in 968 ft. What is this acceleration?
The answer is to be in ft/sec^2 and the answer is .0798.

I'm confused on both and don't know how to get started.
Convert the speeds into feet/sec. Draw a graph of speed vs. time for each runner starting at the moment the fast runner passes. What does the area under the graph represent?

Write out the expression for area under the graph for each runner. What is the condition (in terms of the areas under each of the graphs) for the slow runner to catch the fast one?

AM

ok, I'm needing more assistance.

problem 1: I'm assuming the minimum total velocity is at the peak of its flight. Is this correct? But, I'm still stuck.

problem 2: for the fast runner I have d=14.67(ft/sec)t+4.2 and for the other runner I have d=1/2at^2. I'm not sure on these. That's all I have.

mcr said:
ok, I'm needing more assistance.

problem 1: I'm assuming the minimum total velocity is at the peak of its flight. Is this correct? But, I'm still stuck.
Write out the expression for total velocity in terms of $\vec{v_y} \text{ and } \vec{v_x}$. (hint: you have to add the horizontal and vertical velocities as vectors). How does this total velocity change with time?

problem 2: for the fast runner I have d=14.67(ft/sec)t+4.2
Where does the 4.2 come from?
and for the other runner I have d=1/2at^2. I'm not sure on these. That's all I have.
Not quite. For t<2 what does the graph of velocity vs time look like for the second runner? What is the area under the graph for that part? What does the graph look like for t>=2? What is the area under the graph for that part? Add the areas together. When that is equal to the area under the first graph, what has occurred?

AM

1:The minimum of its velocity is exactly when it is at the peak(you can prove it according to the law of conservation of mechanical energy).Here you may find Vx is V*cosa and Vy is zero,therefore Vmin is V*cosa(V stands for initial velocity).

Thanks, I finally got #1. On to #2. The velocity vs time graph for one runner is going to be a horizontal line. I found it to be y=14.67. The velocity vs time graph for the other runner has two pieces. For t<=2 I have y=12.57, but for t>2 it is a slanted line (y=mx+b), but I don't know how to get the parameters. This is my stumbling part as of right now.

mcr said:
Thanks, I finally got #1. On to #2. The velocity vs time graph for one runner is going to be a horizontal line. I found it to be y=14.67. The velocity vs time graph for the other runner has two pieces. For t<=2 I have y=12.57, but for t>2 it is a slanted line (y=mx+b), but I don't know how to get the parameters. This is my stumbling part as of right now.
What does the slope represent?

Let the slope be a. What is the area under that part of the graph in terms of a?

AM

Andrew Mason said:
What does the slope represent?

Let the slope be a. What is the area under that part of the graph in terms of a?

AM

So the slope is the acceleration, but what about b?

mcr said:
So the slope is the acceleration, but what about b?
Right - slope is acceleration. Don't worry about b. You just need to work out the equation for area under the graph for t>2 (hint: it consists of a triangle and a rectangle - work out the area of each as a function of time t' = t-2)

AM

Andrew Mason said:
What is the condition for minimum total velocity? (hint: does the horizontal velocity change during flight - ignoring air resistance -?)

Convert the speeds into feet/sec. Draw a graph of speed vs. time for each runner starting at the moment the fast runner passes. What does the area under the graph represent?

Write out the expression for area under the graph for each runner. What is the condition (in terms of the areas under each of the graphs) for the slow runner to catch the fast one?

AM

Ok, this is what I got for the area under the curves which represents their distances. For the faster runner I got d=14.7t. For the slower runner I got d=12.6t+1/2a(t-2)^2. So now the two distances are equal so 14.7t=12.6t+1/2a(t-2)^2. Is this correct? I have one equation with two unknowns. So I need to solve for t by coming up with another equation. This is where I am lost. All help is appreciated.

mcr said:
Ok, this is what I got for the area under the curves which represents their distances. For the faster runner I got d=14.7t. For the slower runner I got d=12.6t+1/2a(t-2)^2. So now the two distances are equal so 14.7t=12.6t+1/2a(t-2)^2. Is this correct? I have one equation with two unknowns. So I need to solve for t by coming up with another equation. This is where I am lost. All help is appreciated.
That looks right.

Have another look at the problem. What is the distance covered by the slow runner (after 2 seconds)?

It might be easier to break it down:

Distance traveled by slow runner in first two seconds = 12.6 x 2 = 25.2 feet
Distance traveled by slow runner after 2 seconds = 12.6(t-2) + .5a(t-2)^2

So the equation is: 14.7t = 25.2 + 12.6(t-2) + .5a(t-2)^2AM

Andrew Mason said:
That looks right.

Have another look at the problem. What is the distance covered by the slow runner (after 2 seconds)?

It might be easier to break it down:

Distance traveled by slow runner in first two seconds = 12.6 x 2 = 25.2 feet
Distance traveled by slow runner after 2 seconds = 12.6(t-2) + .5a(t-2)^2

So the equation is: 14.7t = 25.2 + 12.6(t-2) + .5a(t-2)^2

AM

So I proceeded to use the equation: 12.6(t-2)+.5a(t-2)^2=198 ft. and solved for (t-2) in terms of a. I then substituted back into the equation and I get a=.0663 ft/sec^2. This is not the answer, so I must have done something incorrect. Does anything stand out that is wrong? Thanks.

mcr said:
So I proceeded to use the equation: 12.6(t-2)+.5a(t-2)^2=198 ft. and solved for (t-2) in terms of a. I then substituted back into the equation and I get a=.0663 ft/sec^2. This is not the answer, so I must have done something incorrect. Does anything stand out that is wrong? Thanks.
?? Where does the 198 feet come from? Your statement says 968 feet. Your answer is correct if the question says 968 feet. Better check the question again.

AM

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## 1. How can I improve my problem-solving skills in physics?

Improving problem-solving skills in physics requires practice and patience. Start by understanding the basic concepts and principles of physics, and then work on solving different types of practice problems. It is also beneficial to break down complex problems into smaller, more manageable steps.

## 2. What are some effective strategies for solving physics practice problems?

Some effective strategies for solving physics practice problems include identifying and understanding the given information, drawing diagrams or visual representations, and using relevant equations and formulas. It is also helpful to check units and significant figures throughout the problem solving process.

## 3. How do I know which equation to use when solving a physics problem?

The equation to use when solving a physics problem depends on the given information and what is being asked in the problem. It is important to read the problem carefully and identify the known and unknown variables. Then, use the relevant equations that relate these variables and solve for the unknown quantity.

## 4. What should I do if I am stuck on a difficult physics practice problem?

If you are stuck on a difficult physics practice problem, take a step back and review the concepts and principles involved. It may also be helpful to look at similar problems or seek assistance from a teacher or tutor. Additionally, breaking the problem into smaller parts or approaching it from a different angle can also aid in finding a solution.

## 5. How can I check if my answer to a physics practice problem is correct?

To check if your answer to a physics practice problem is correct, you can use the given information and the equations used to solve the problem to verify your answer. It is also helpful to check the units and significant figures to ensure accuracy. If possible, it is always beneficial to have someone else review your solution for any errors or mistakes.

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