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A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3/2 km in 3.5 min. How fast (in m/s) is the car moving after this time?

(the equation that my teacher taught us is d= ((vi + vf)/2) x t (where d = distance, vi = initial velocity, vf = final velocity, t = time), so if you could use that it would be greatly appreciated.)

(please show all of your work or add a comprehensive explaination)

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StudiousStud3nt

Answer:

7.9m/s

Explanation:

we need to convert the given values to metres and seconds

3/2km = 1500m and 3.5min = 210s

Using the first key equation of accelerated motion:

d = ((vi+vf)/2) x t

1500m = ((6.4+vf)/2) x 210 (plug in values)

3000m = (6.4 + vf) x 210 (get rid of the denominator by multiplying both sides by 2)

14.29 = 6.4 + vf (divide both sides to get rid of time)

vf = 7.885 m/s (subtract 6.4 from the right side to isolate vf)

2 significant digit answer is 7.9m/s

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