Question relates to Degree and some other simple question.

  • Thread starter Thread starter louis676
  • Start date Start date
  • Tags Tags
    Degree
AI Thread Summary
Marcia's car took 3.1 seconds to stop after the light turned red, traveling approximately 29.76 meters before coming to a halt. For the second scenario, Ali's ball was thrown at a 60-degree angle with an initial speed of 25 m/s, but further calculations are needed to determine its height from Ali. In the third scenario, Maxim's car traveled 5 meters in 2.5 seconds while accelerating at 5.2 m/s². The discussions highlight confusion around using the correct equations and understanding the concepts of acceleration and motion. Clarification on the negative acceleration and visual aids like velocity-time graphs were suggested for better understanding.
louis676
Messages
26
Reaction score
0
1. Marcia was driving her car at the constant speed of 19.2 m/s when she saw the light turn red. It took her .13 sec to hit the brakes, but then the car slowed at the constant rate of 6.2 m/s2 until it stopped. How much time did it take for the car to stop moving after the light turned red.? How far did the car travel until it stopped?
Did I plug in the right number for each symbol?


I am confuse of which equation to use and which question should I put .13s to.

First question:

V0 = 19.2 m/s
A = -6.2 m/s 2
Δt = ?

Δx=?

Δx = V0Δt + .5a(Δt)2

Second question:
V0 = 19.2 m/s
A = -6.2 m/s 2
Δt = .13 s

Δx=?

Δx = V0Δt + .5a(Δt)2

2. Standing on a flat roof of a 21 m high building, Ali threw a ball up at a 60 degree angle to the roof top. The speed of the ball was 25 m/s. When the ball left Ali’s outstretched hand the ball was 3.1 meters from the roof top. The ball eventually landed on the roof. High far was the ball from Ali?

I never work on question related to degree, so I am pretty lost on this one.
Vo = 25m/s
A = -9.8 m/s 2
Vf = 0 m/s
Δx = ?







3. Maxim was driving his car at 15 m/s when he pressed the gas pedal down to accelerate at 5.2 m/s2. How far did he travel in the next 2.5 seconds?

I think this is correct.
Δx = ?
V0 = 15 m/s
Δt = 2.5 s
A = -5.2 m/s 2

Δx = V0Δt + .5a(Δt)2

Δx = 15 m/s (2.5 s) – 5.2 m/s 2 (2.5 s 2)

Δx = 37.5 m – 32.5 m

Δx = 5m
 
Physics news on Phys.org
Bump!
 
1. Draw a velocity - time graph.
2. Not sure how the rooftop and roof configuration.
3. Why the acceleration is negative? Draw a velocity- time graph here too.
 
Last edited:
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top