Velocity and Acceleration Equational Problem ( making sense)

Click For Summary

Homework Help Overview

The discussion revolves around a kinematics problem involving a man running to catch a bus that starts moving with acceleration. The original poster is trying to determine the position of the man as a function of time, given his initial distance from the bus and his constant speed.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the velocity equation but expresses confusion regarding the use of speed versus acceleration. They question the relationship between the man's constant speed and the bus's acceleration.
  • Another participant suggests using a different equation for displacement, indicating that the acceleration of the man is zero, which prompts further reflection from the original poster.

Discussion Status

Participants are engaging in clarifying the correct equations to use for the problem. One participant has provided guidance on the appropriate equation for displacement, which has helped the original poster gain clarity. However, there is still an exploration of the underlying concepts regarding speed and acceleration.

Contextual Notes

The original poster expresses a desire to understand foundational concepts better, indicating a focus on learning rather than just finding the answer. There is an acknowledgment of confusion regarding the application of kinematic equations in this context.

zippeh
Messages
12
Reaction score
0

Homework Statement



A man is running at speed c (much less than the speed of light) to catch a bus already at a stop. At t=0, when he is a distance b from the door to the bus, the bus starts moving with the positive acceleration a. Use a coordinate system with x=0 at the door of the stopped bus.

Part A:
What is xman(t), the position of the man as a function of time?
Answer symbolically in terms of the variables b, c, and t.

Homework Equations



v(t) = V0+at



The Attempt at a Solution



I know that I need to use this equation:

v(t) = V0+at

But I am confused because the answer is: xman(t) = -b+ct. This doesn't make sense to me because in order to use this equation, c would need to be the acceleration, not the velocity. I tried reviewing the book and it really doesn't help. I know that the mans speed is constant, but with an acceleration c, wouldn't his velocity increase in time?

Sorry if this is a simple question, I just am trying to understand the foundations so that I can better apply myself later in the course. Thank you for your time.
 
Physics news on Phys.org
zippeh said:

Homework Statement



A man is running at speed c (much less than the speed of light) to catch a bus already at a stop. At t=0, when he is a distance b from the door to the bus, the bus starts moving with the positive acceleration a. Use a coordinate system with x=0 at the door of the stopped bus.

Part A:
What is xman(t), the position of the man as a function of time?
Answer symbolically in terms of the variables b, c, and t.

Homework Equations



v(t) = V0+at

The Attempt at a Solution



I know that I need to use this equation:

v(t) = V0+at

But I am confused because the answer is: xman(t) = -b+ct. This doesn't make sense to me because in order to use this equation, c would need to be the acceleration, not the velocity. I tried reviewing the book and it really doesn't help. I know that the mans speed is constant, but with an acceleration c, wouldn't his velocity increase in time?

Sorry if this is a simple question, I just am trying to understand the foundations so that I can better apply myself later in the course. Thank you for your time.

Welcome to PF!

If you are asked to find displacement , why should you be using equation for finding velocity .

You need to use the equation x(t) = x0 + ut+(1/2)at2 .

Here x0 = -b , u=c , a=0 .

Does that make sense ?
 
Oh wow... thanks!

I was looking through the equations that I know, and I couldn't arrive at the answer with anything that I had (so I thought). I immediately looked for an equation that looked similar to the answer, which was what I gave above. Thanks for making it clear, I didn't really ever think to set accel. to 0, which was what was made me stuck. I appreciate you helping me! :)
 
You are welcome :)
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
22
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K