Uniformly accelerated motion issue

In summary: So, you're saying that the car was traveling at ##5.6m/s## at the middle of the third second?Exactly!So, you're saying that the car was traveling at ##5.6m/s## at the middle of the third second?
  • #1
greg_rack
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Homework Statement
A car is travelling along a straight road with constant acceleration. It passes a road sign. It travels 12.2 m in the 3rd second and 14.4 m in the 4th second after passing the road sign. What was the speed of the car as it passed the road sign?
Result: [6.70m/s]
Relevant Equations
UAM law of time
a=Δv/Δt
v=Δx/Δt
I went quite confidently on this one, at least at the beginning...
I found the instant velocity on the 3rd second(v1=12.2m/1s) and the one on the 4th(v2=14.4m/1s), and subsequently the acceleration(a=2.2/1m/s*s).
Since the acceleration is constant, using the law "v=v0+aΔt", I have found v0 which in my case results as 5.6m/s as opposed to the 6.7m/s it should.
What am I getting wrong?
 
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  • #2
greg_rack said:
Homework Statement:: A car is traveling along a straight road with constant acceleration. It passes a road sign. It travels 12.2 m in the 3rd second and 14.4 m in the 4th second after passing the road sign. What was the speed of the car as it passed the road sign?
Result: [6.70m/s]
Relevant Equations:: UAM law of time
a=Δv/Δt
v=Δx/Δt

I went quite confidently on this one, at least at the beginning...
I found the instant velocity on the 3rd second(v1=12.2m/1s) and the one on the 4th(v2=14.4m/1s), and subsequently the acceleration(a=2.2/1m/s*s).
Since the acceleration is constant, using the law "v=v0+aΔt", I have found v0 which in my case results as 5.6m/s as opposed to the 6.7m/s it should.
What am I getting wrong?
At what time in the third second did you calculate the velocity? At the beginning, middle or end of that interval?
 
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  • #3
PeroK said:
At what time in the third second did you calculate the velocity? At the beginning, middle or end of that interval?
I supposed it ran 12.2m in one second, from the beginning to the end of the 3rd second interval.
 
  • #4
greg_rack said:
I supposed it ran 12.2m in one second, from the beginning to the end of the 3rd second interval.
The question tells you that. But, you calculated a velocity. At what time does that velocity apply?
 
  • #5
PeroK said:
The question tells you that. But, you calculated a velocity. At what time does that velocity apply?
That applies at the end of the interval, I'd say
 
  • #6
greg_rack said:
That applies at the end of the interval, I'd say
Are you saying that the car was traveling at ##12.2m/s## at the end of the third second?
 
  • #7
PeroK said:
Are you saying that the car was traveling at ##12.2m/s## at the end of the third second?
Oh, well, since it's a "mean" velocity, shouldn't it apply to the middle of the third second?
 
  • #8
greg_rack said:
Oh, well, since it's a "mean" velocity, shouldn't it apply to the middle of the third second?
Exactly!
 
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Related to Uniformly accelerated motion issue

1. What is uniformly accelerated motion?

Uniformly accelerated motion is a type of motion in which an object moves with a constant acceleration. This means that the object's velocity increases by the same amount in each unit of time.

2. How is uniformly accelerated motion different from uniformly constant motion?

Uniformly constant motion is when an object moves with a constant velocity, meaning it does not change in speed or direction. In uniformly accelerated motion, the object's velocity is changing at a constant rate.

3. What is the equation for calculating uniformly accelerated motion?

The equation for calculating uniformly accelerated motion is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

4. How does the acceleration of an object affect its motion?

The acceleration of an object determines how quickly its velocity changes. If the acceleration is positive, the object's velocity will increase; if it is negative, the velocity will decrease. The greater the acceleration, the faster the change in velocity.

5. Can an object have both a constant velocity and a constant acceleration?

No, an object cannot have both a constant velocity and a constant acceleration. This is because a constant acceleration means the object's velocity is changing, while a constant velocity means the object's velocity is not changing. These two concepts are contradictory.

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