- #1

greg_rack

Gold Member

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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?

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?