d = Vo + 0.5at^2.
d=9.8 + 0.5a(1.4)^2=-8.1 -1.38 = -9.48,
9.8 + 0.98a = -9.48,
0.98a = -9.48 - 9.8 = -19.28,
a = -19.7cm/s^2.
A body moving with uniform acceleration has a velocity of 9.8 cm/s when its x coordinate is 1.38 cm. If its x coordinate 1.4 s later is -8.1 cm, what is the magnitude of its acceleration? Answer in units of cm/s2
2 answers
correction: d = Vo*t + 0.5at^2.
9.8*1.4 + 0.5*a(1.4)^2 = -8.1 -1.38,
13.72 + 0.98a = -9.48,
0.98a = -9.48 -13.72 = -23.2,
a = -23.7cm/s^2.
9.8*1.4 + 0.5*a(1.4)^2 = -8.1 -1.38,
13.72 + 0.98a = -9.48,
0.98a = -9.48 -13.72 = -23.2,
a = -23.7cm/s^2.