In all kinematics, 't' represents time.
(vf-vi)/ t is ?
vf=vi+at² simplifying making vi=0, v=at²t²=v/at=√v/atime equals square root of velocity divided by acceleration (or gravity)
Speed=distance/time Velocity=displacement/time Acceleration=VI-VF/T or or OR S=d/t V=d/t a=vi-vf/t
(t - 10)t=17 = 7(t - 10)t=50 = 40
acceleration cannot be calculated from these values alone unless one makes a few assumptions: Vf=final velocity Vi=initial velocity a=acceleration d=displacement t=time assume Vi=0 (Vf-Vi)/t=a Vf=at+Vi Vf**2=Vi**2+2ad (at)**2=2ad aatt=2ad att=2d a=2d/t**2
(vf-vi)/ t is ?
Yes recalling the first equation of motion ie Vf = Vi + at Here Vf is final velocity and Vi is the initial velocity. a the acceleration and t is the time Now taking at on the other side ie left side we get Vf - at = Vi This is what mentioned here.
vf=vi+at² simplifying making vi=0, v=at²t²=v/at=√v/atime equals square root of velocity divided by acceleration (or gravity)
If A = T and T = U then A = U.
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T. D. Dasan Std. VI B was created on 2010-04-09.
Speed=distance/time Velocity=displacement/time Acceleration=VI-VF/T or or OR S=d/t V=d/t a=vi-vf/t
s is directly proportional to t
(t - 10)t=17 = 7(t - 10)t=50 = 40
50
d=vi*t+.5(9.81)(t)^2
Divide the change in position, (total distance covered) by the time it took. Xf = xi +at a = xf-xi / t That is the definition of velocity, not acceleration. Acceleration is rate of change of velocity. (vfinal - vinitial)/t for constant acceleration so vf equals vi + at. Or a equals dV/dt otherwise.