Beginner Physics Help - Questions & Solutions

[omelfata]

Hello, I just started a Physics course last week and unfortunatly I am stuck with a teacher that can't teach So , I have come across a few questions that I need help in.


1. A car traveling at 90km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m . What was the average acceleration od th driver during the collision?

2. A stone dropped from the top of a cliff. It is seen to hit the ground below after 3.50seconds. How high is the cliff?

3. A heliocopter is ascending vertically with a speed of 5.50m/s . At a hight of 105m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

Thank you in advance.

 

[Kurdt]

Welcome to PF! Unfortunately the rules of the forum are such that we can not help you unless you've shown some attempt at doing the questions yourself. This is for your own benefit as the learning process would be stunted by people just handing out answers. Please tell us how you are stuck with the questions (i.e. how far have you gotten yourself and what stage are you stuck at).

 

[kyle8921]

Hello, it's great that you are taking a physics course! Physics can be a challenging subject, but with some practice and understanding, it can also be very rewarding. I would be happy to help you with these questions.

1. To find the average acceleration of the driver, we can use the formula a = (vf - vi)/t, where vf is the final velocity (in this case, 0 m/s), vi is the initial velocity (90 km/h converted to m/s), and t is the time it took for the driver to come to rest (0.80 m converted to seconds). Plugging in the values, we get a = (0 m/s - 25 m/s)/0.80 s = -31.25 m/s^2. Since the acceleration is negative, this means the car was decelerating during the collision.

2. To find the height of the cliff, we can use the formula h = 1/2gt^2, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time it took for the stone to hit the ground (3.50 s). Plugging in the values, we get h = 1/2 * 9.8 m/s^2 * (3.50 s)^2 = 60.9 m. So the cliff is 60.9 meters high.

3. To find the time it takes for the package to reach the ground, we can use the formula h = vi*t + 1/2at^2, where h is the height (105 m), vi is the initial velocity (5.50 m/s), and a is the acceleration due to gravity (-9.8 m/s^2). We can rearrange this equation to solve for t, which gives us t = (-vi ± √(vi^2 - 4(1/2)a(105)))/(-9.8). Plugging in the values, we get t = (-5.50 ± √(5.50^2 - 4(1/2)(-9.8)(105)))/(-9.8) = 5.90 seconds. So it takes 5.90 seconds for the package to reach the ground.

I hope this helps! If you have any other questions or need further clarification, please don't hesitate to ask. Remember to keep practicing and don't be afraid