v = u + at where u = starting velocity, v = final velocity, a = acceleration, t = time. Here u = 0 so v = at ie a = v/t Now, v = 5 m/s (what is mk?) and t = 5 s So a = (5 m/s) / 5 s = 1 m/s2
v2- u 2 = 2assince, S (Distance) = Average speed x TimeS = U+V / 2 * TS = U+V / 2 * V - U / A {since T = V -U / A}S = V2 - U2 / 2A2AS = V2 - U2OR V2 - U2 = AsHence, Derived.
For a relation, $, to be called a partial ordering on a set, S, the following three properties must be met:1) If T is any subset of S, then T $ T.2) If T and U are any two subsets of S that meet the condition T $ U as well as the condition U $ T, then T = U.3) If T, U, and V are any three subsets of S that meet the condition T $ U as well as the condition U $ V, then T $ V.For the relation, $, to be called a total ordering on the set, S, the following statement must hold in addition to the previous three:If T and U are any two subsets of S, then either T $ U or U $ T.This final property is called totality.For an example of a partial ordering relation, see the related link on "less than or equal to."Also, see the corresponding related link for the definition of "relation."
Where a = (v-u)/t a is acceleration, v is final velocity u is initial velocity t is time so, u=v-at
acceleration(a) = (final velocity(v) - Initial velocity(u)) / time (s) Algebraically a = (v - u) / t Where 'v' & 'u' are measured in metres per second ( m/s) or ms^-1 And 't' is the time in seconds measured is 's' Hence a(ms^-2) = v(m/s) - u(m/s)) / t(s) And example is a car starting from rest up to 44 m/s ( 30 mph) in 10 seconds. a = (44 - 0 ) / 10 a = 44/10 a = 4.4 ms^-2. NB Earth's gravitational acceleration(g) is approximately 10 ms^-2.
Suppose you have the fractions p/q and r/s. Let the LCM of q and s be t.Then t is a multiple of q as well as of s so let t= q*u and t = s*v Then p/q = (p*u)/(q*u) = (p*u)/t and r/s = (r*v)/(s*v) = (r*v)/t have the same denominators.
v = u + at where u = starting velocity, v = final velocity, a = acceleration, t = time. Here u = 0 so v = at ie a = v/t Now, v = 5 m/s (what is mk?) and t = 5 s So a = (5 m/s) / 5 s = 1 m/s2
If: v = u+at Then: -u = -v+at or u = v-at (by dividing all terms by -1) a = (v-u)/t t = (v-u)/a
the 4 fundamental kinematic equations of motion for constant acceleration (suggest you commit these to memory): s = ut + Β½at^2 β¦. (1) v^2 = u^2 + 2as β¦. (2) v = u + at β¦. (3) s = (u + v)t/2 β¦. (4) where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time. In this case, we know u = 10m/s, v = 50m/s, t = 10s and we want to find a, so we use equation (3) v = u + at 50 = 10 + 10t, so t = (50 β 10)/10 = 4m/s^2
v = 2s/t - u where u=initial velocity, v=final velocity, s = distance and t = time
O b v i o u s l y y o u s n i t c h
v2- u 2 = 2assince, S (Distance) = Average speed x TimeS = U+V / 2 * TS = U+V / 2 * V - U / A {since T = V -U / A}S = V2 - U2 / 2A2AS = V2 - U2OR V2 - U2 = AsHence, Derived.
Well, V=s/tso, V/t = (s/t)/tor V/t = s/t * 1/tso, V/t= s/(t^2). It is equal to acceleration.
why is this a question? i t s o b v i o u s
change in velocity (v) = acceleration (a) x time (t); distance s = 1/2 a times t squared; solve for time and substitute; find a = v squared /(2s)
v
S=vt-16t2 solve for v is what I will assume you mean. first pull out the t S=t(v-16t) then devide by t S/t=v-16t Then add 16t to both sides S/t + 16t = v This can also be written as (S+16t2)/t = v