d= vi*t+1/2 a t^2
60=6Vi+18a
but vf=vi+at
15=vi+6a
so you have two equations, two unknowns, solve. Multiply the second equation by 3, and subtract it frm the first is easy way.
A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passed the second point was 15.0 m/s. (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph x versus t and v versus t for the car, from rest (t = 0).
3 answers
Vi = 5m/s
a = 1.666
a = 1.666
What about part c and d????
1 answer