Physic (kinematics) — Displacement of an accelerating car

In summary, the displacement of the car at t=15s can be found by using the equation s = si + vit + ½(a)(Δt)^2, which takes into account the non-linear relationship between distance and time due to acceleration. The correct answer is 2.3 x 10^2 m, which should be expressed to two significant figures.
  • #1
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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  • #2
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)
 
  • #3
You used the final speed for 15 sec.
You might ask "What was the average speed during those 15 sec.?"
 
  • #4
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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  • #5
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?
 
  • #6
Thank you for your help :)
 
  • #7
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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  • #8
Thank you so much for helping
 

Related to Physic (kinematics) — Displacement of an accelerating car

What is displacement?

Displacement is a vector quantity that measures the change in position of an object from its initial position to its final position.

How is displacement different from distance?

Distance is a scalar quantity that measures the total length of the path traveled by an object, while displacement is a vector quantity that only takes into account the change in position.

What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity and is represented by the symbol "a".

How is acceleration related to displacement?

Acceleration is directly related to displacement through the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

How can we calculate displacement for an accelerating car?

To calculate displacement for an accelerating car, we can use the equation d = vit + 1/2at2, where d is displacement, vi is initial velocity, a is acceleration, and t is time.

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