d = 1/2 a t^2 + v0 t + d0
-5.00 = 1/2 a 2.95^2 + 13.0(2.95) + 2.73
An object moving with uniform acceleration has a velocity of 13.0 cm/s in the positive x-direction when its x-coordinate is 2.73 cm. If its x-coordinate 2.95 s later is −5.00 cm, what is its acceleration? cm/s^2
Express the position of the object in one-dimensional motion at constant acceleration as a function of time. Use the value of the expression at the two times to find the acceleration.
2 answers
V1 = 13cm/s.
V2 = d/t = (-5-2.73)/2.95 = -2.62cm/s.
Acceleration = the change in velocity / Time to make the change:
a = (V2-V1)/t = (-2.62-13)/2.95 =
V2 = d/t = (-5-2.73)/2.95 = -2.62cm/s.
Acceleration = the change in velocity / Time to make the change:
a = (V2-V1)/t = (-2.62-13)/2.95 =
2 answers