3
a=s/t, and s=d/t, so if we substitute... a = (d/t)/t --> a = d/t2 You must know both the acceleration and time in order to solve for the distance travelled.
You can solve for a one-time constant by using the formula t = RC. Read the math problem you are given carefully to determine what values to plug into the equation.
You can't "solve" since this isn't "=" to anything.
If the initial velocity was zero then final velocity V = 2*S / t where S = distance covered and t = time it took Acceleration a = 2*S / t2 or a = (V - V0) / t where (V - V0) is change in velocity.
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
you take a s**t
P=s r t , so, s= P/(st)
SOLVE : 21t + t=
4 solve t
S = 3 st + 3t = 6 st = 6 - 3t s = (6 - 3t)/t = 6/t - 3
a=s/t, and s=d/t, so if we substitute... a = (d/t)/t --> a = d/t2 You must know both the acceleration and time in order to solve for the distance travelled.
Given T = R + RS Lateral inversion makes it to be R + RS = T Taking R as common factor, we get R(1+S) = T Now dividing by (1+S) both sides, R = T / (1+S) Hence the solution R = T/(1+S)
If you have an expression with many variables, you can often solve for a different one. One classic example is rate x time = distance. We use the variable rxt=d Now this is solved for d, but say you want to solve for t. You divide both sides by r and t=d/r and it is solved for t.
In order to solve for t, we need to get it alone on one side.10 = 2t5 = t divide both ides by 2
T. G. Poole has written: 'Using simulation to solve problems' -- subject(s): Digital computer simulation
let us assume that t is time, d is distance, and s is speed. You could therefore use the following equation to solve this kind of problem: s=d/t Using the above example you could solve for the amount of time it would take to complete this trip by substituting values in the above equation as shown: s=d/t ts=d t=d/s t=(375 mi)/(50 mph)