Physic (kinematics) — Displacement of an accelerating car

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Homework Help Overview

The discussion revolves around a kinematics problem involving the displacement of a car accelerating from rest at a rate of 2.0 m/s² over a time period of 15 seconds. Participants are exploring the correct approach to calculate the displacement, given that the original poster's calculations led to an incorrect result.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's method of using the final speed to calculate displacement and question whether the formula s = vt is applicable in this scenario. Some suggest considering average speed during the acceleration period and refer to the appropriate kinematic equations.

Discussion Status

There is an ongoing exploration of the correct kinematic equations to use, with participants providing hints and questioning the assumptions made in the original calculations. Some guidance has been offered regarding the need to consider acceleration and average speed, but no consensus has been reached on the final approach.

Contextual Notes

Participants note that the original poster's calculations yield a displacement of 450 m, which is inconsistent with the expected answer of 230 m. There is also mention of the need for significant figures in the final answer, indicating a focus on precision in the problem setup.

Liszzy
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Question: A car accelerates from rest at a rate of 2.0m/s^2 [N]. What is the displacement of the car at t=15s?

I tried
(2m/s^2 [N])(15s)=30m/s [N]
(30m/s [N])(15s)
=450m [N] but guess wrong :(

*Because the answer is 230m [N]

How to solve this? anyone
 
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Liszzy said:
Question: A car accelerates from rest at a rate of 2.0m/s^2 [N]. What is the displacement of the car at t=15s?

I tried
(2m/s^2 [N])(15s)=30m/s [N]
(30m/s [N])(15s)
=450m [N] but guess wrong :(

*Because the answer is 230m [N]

How to solve this? anyone

So what you did was the following. First: ##v=at##, then: ##s=vt##.

In universities, when they teach this subject. They usually derive a set of equations before doing such a problem. But, since you did what you did, for now, where do you think your mistake is? Is ##s=vt## all the time?

Hint: we know that the actual equation is ##v=\frac{ds}{dt}## ||| ##a=\frac{dv}{dt}## (the second won't be really necessary, but you can use it)
 
You used the final speed for 15 sec.
You might ask "What was the average speed during those 15 sec.?"
 
Liszzy said:
Question: A car accelerates from rest at a rate of 2.0m/s^2 [N]. What is the displacement of the car at t=15s?

I tried
(2m/s^2 [N])(15s)=30m/s [N]
(30m/s [N])(15s)
=450m [N] but guess wrong :(

*Because the answer is 230m [N]

How to solve this? anyone
I think that your line of reasoning implies that the displacement is linear to time. However, since there exists an acceleration a, then the distance traveled at one moment in time is different than the distance traveled in the next moment. There are three equations you should think of when you are considering kinematics. For this problem, you should consider this equation.
s = si + vit + ½(a)(Δt)2
 
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Liszzy said:
Question: A car accelerates from rest at a rate of 2.0m/s^2 [N]. What is the displacement of the car at t=15s?

I tried
(2m/s^2 [N])(15s)=30m/s [N]
(30m/s [N])(15s)
=450m [N] but guess wrong :(

*Because the answer is 230m [N]

How to solve this? anyone

Do you have a textbook? Does it solve similar problems?
 
Thank you for your help :)
 
Liszzy said:
Question: A car accelerates from rest at a rate of 2.0m/s^2 [N]. What is the displacement of the car at t=15s?

I tried
(2m/s^2 [N])(15s)=30m/s [N]
(30m/s [N])(15s)
=450m [N] but guess wrong :(

*Because the answer is 230m [N]

How to solve this? anyone
The answer should not be 230m - it should be 2.3 x 10^2 m
The numeric answer is 225, but, should be expressed to 2 significant figures due to the precision in the supplied information.
 
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Thank you so much for helping
 

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