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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
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