What is the speed of the sports car at the second marker?

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The speed of a sports car at the second marker, after accelerating at 1.8 m/s² over a distance of 120 meters in 4.1 seconds, is calculated to be 44.3 m/s. The initial speed (vi) was determined to be 14.2 m/s using the equation d = vi(t) + 1/2(a)t². The final speed (vf) was derived from the average speed formula d = 1/2(vi + vf)t, confirming that vf equals 44.3 m/s. The calculations were verified, but one participant suggested a re-evaluation of the numbers.

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A sports car, picking up speed, passes between two markers in a time of 4.1s. The markers are separated by 120m. All the while, the car accelerates at 1.8m/s^2. What is its speed at the second marker?

Here is what i did:

d = vi(t)+1/2(a)t^2
120=vi(4.1)+1/2(1.8)(4.1)^2
29.3=vi+15.1
vi=14.2m/s

d=1/2(vi+vf)t
120=1/2(14.2+vf)(4.1)
vf=44.3m/s
vf=25.2m/s
 
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miamiheat5 said:
A sports car, picking up speed, passes between two markers in a time of 4.1s. The markers are separated by 120m. All the while, the car accelerates at 1.8m/s^2. What is its speed at the second marker?

Here is what i did:

d = vi(t)+1/2(a)t^2
120=vi(4.1)+1/2(1.8)(4.1)^2
29.3=vi+15.1
vi=14.2m/s

d=1/2(vi+vf)t
120=1/2(14.2+vf)(4.1)
vf=44.3m/s
vf=25.2m/s

Your calculation is wrong. Check the numbers once again.
 

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