Question

a car traveling at 7.0 m/s accelerates uni formally at 2.5 m/s^2 to reach a speed of 12.0 m/s how long does it take for the acceleration to occur

Answer 1

2

vf = vi +at
12= 7 +2.5t
12-7 = 2.5t
5 = 2.5t
5 / 2.5 = t
2 = t

Answer 2

what formula are you using man because am lost

Answer 3

\[v=vi + a \Delta t\]

Answer 4

v is the final velocity = 12.0 m/s
vi is the initial velocity = 7.0 m/s
a is the acceleration = 2.5 m/s2
and "triangle t" (which is pronounced delta t) is the change in time or how long it takes

Answer 5

thank you so ,much i got a way different equation in my book

Answer 6

turner's treadmill runs with a a velocity of -1.2m/s and speeds up at regular intervals during a half hour work out . after 25 min the treadmill has a velocity of -6.5 m/s what is the average acceleration of the treadmill

Answer 7

i got about 1.7

Answer 8

There is a way to make the formulas if you know some calculus. Starts with acceleration, which is a change of velocity divided by the change of time. \[a=\Delta V /\Delta t\]rearranged as \[a \Delta t = \Delta V\] and if INTEGRATED becomes \[at + c = V\] c is a constant valued as the velocity present at the measurement of THE TIME, so it becomes \[at + V(present) = V(t)\] Velocity is the change of distance divided by the change of time. \[V = \Delta X / \Delta t\] rearranged as \[V \Delta t = \Delta X\] which is same as \[[ at + V(t)] \Delta t = \Delta X\]and if INTEGRATED becomes \[[at ^{2}/2] + V(present) t + c = X(t)\] now c is the distance at which the mobile is present at the measurement of THE TIME\[X(t) = X(present) + V(present) t + [at ^{2 }/2]\] many times V(present) =0 or X(present) = 0 or a = 0