Physic (kinematics) — Displacement of an accelerating car

AI Thread Summary
A car accelerates from rest at 2.0 m/s², and the displacement after 15 seconds is calculated incorrectly as 450m. The correct displacement is 225m, which should be expressed as 2.3 x 10² m to reflect significant figures. The mistake arises from using the final speed for the entire duration instead of calculating average speed during acceleration. The appropriate kinematic equation to use is s = si + vit + ½(a)(Δt)², which accounts for the changing speed. Understanding the relationship between acceleration, velocity, and displacement is crucial for solving such problems accurately.
Liszzy
Messages
6
Reaction score
3
Member has been warned not to delete the template.
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
 
Last edited by a moderator:
  • Like
Likes Mason Smith
Physics news on Phys.org
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
 
  • Like
Likes Liszzy and wormbread
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.
 
  • Like
Likes Liszzy
Thank you so much for helping
 
Back
Top