Time taken to accelerate and travel 100m How do i work out how long it would take to accelerate from 0 and travel 100 m  with a terminal speed of 110km per hr
can not seem to find formula to solve this without knowing acceleration 
please assist
thank you 
 A: Let's assume constant acceleration, since some assumption about acceleration is necessary to get an answer to your problem.
If you expect to do many constant acceleration problems in the future, it would be worthwhile for you to memorize the equations of motion. These equations are easily derived from calculus and are worth memorizing. The equation for constant acceleration that does not use the acceleration is
$$\Delta x=\frac 12 (v_i+v_t)\Delta t$$
where $\Delta x$ is the displacement during the time period, $v_i$ is the initial velocity, $v_f$ is the final velocity, and $\Delta t$ is the time interval. (Let me know if you want to see a derivation of this equation.) In your case, you want to solve for the time, so
$$\begin{align}
\Delta t &= \frac{2\Delta x}{v_i+v_t} \\[2 ex]
 &= \frac{2\cdot 0.100\text{ km}}{0\text{ km/hr}+110\text{ km/hr}} \\[2 ex]
 &= 0.00182\text{ hr} \\[2 ex]
 &= 6.55\text{ s}
\end{align}$$
Note that I followed the usual high-school rules of precision and assumed that your values of $100$ and $110$ had three significant digits / zero decimal places of precision. Also notice the two changes in units, from $100$ m to $0.100$ km and $0.00182$ hours to $6.55$ seconds. Let me know if you are unsure on how to do those unit changes.
A: If the initial velocity is 0, then we have the following equations:
$$
v = at (1)\\ 
x = \frac { a }{ 2 } { t }^{ 2 } (2)
$$
Now divide $(2)$ by $(1)$ to get:
$$
\frac { x }{ v } \quad =\quad \frac { \frac { a }{ 2 } { t }^{ 2 } }{ at } \\ \frac { x }{ v } \quad =\quad \frac { t }{ 2 } \\ t\quad =\quad \frac { 2x }{ v } 
$$
Now we have the terminal velocity ${ v }_{ f }\quad =\quad 110\quad km/h\quad =\quad \frac { 110 }{ 3.6 } m/s$ and $d = 100m$, so our final result is:
$$
t\quad =\quad \frac { 2d }{ { v }_{ f } } \\ t\quad =\quad \frac { 200\quad m }{ \frac { 110 }{ 3.6 } m/s } \\ t\quad =\quad 6.54\quad s
$$
A: If you want to solve this problem, with or without constant acceleration, use the following formulas
$$\begin{align}
\Delta x &= \int \limits_0^v \frac{v}{a(v)}\,{\rm d}v & \Delta t &= \int \limits_0^v \frac{1}{a(v)}\,{\rm d}v
\end{align}$$
So if this is a car, with acceleration (ignore air drag) under power $P$
$$ a(v) = \frac{P}{m v}$$
$$\begin{align}
\Delta x &= \int \limits_0^v \frac{m v^2}{P}\,{\rm d}v & \Delta t &= \int \limits_0^v \frac{ m v}{P}\,{\rm d}v \\ 
\Delta x & = \frac{m v^3}{3 P} & \Delta t &= \frac{m v^2}{2 P}
\end{align}$$
Now fit $v=110 \mbox{ km/h} = 30.55 \mbox{ m/s}$ and $\Delta x = 100 \mbox{ m}$ to get
$$ \begin{align} \frac{P}{m} & = \frac{v^3}{3 x} = 95.093 \mbox{ W/kg} & \Delta t & = \frac{3 x}{2 v} = 4.909\mbox{ sec} \end{align}$$
A: Let $a$ be the acceleration, $u$ be the initial velocity & $v=110\ km/ hr=\frac{110\times 1000}{3600}=\frac{275}{9}\ m/sec $ be the terminal (final) velocity achieved in time $t$ by travelling a distance $s=100\ m$. 
Now, applying first equation of motion as follows $$v=u+at$$  since, the motion starts from rest hence substituting $u=0$, $v=\frac{275}{9}$ we get
$$\frac{275}{9}=0+at\implies a=\frac{275}{9t}$$ 
Applying third equation of motion as follows $$v^2=u^2+2as$$ substituting $u=0$, $v=\frac{275}{9}$ & $s=100$, we get 
$$\left(\frac{275}{9}\right)^2=(0)^2+2\left(\frac{275}{9t}\right)(100)$$ $$\implies t=\frac{200\times 9}{275}$$$$=\frac{72}{11}\ sec$$ Hence we have 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{time taken, t}=\frac{72}{11}\approx 6.54 \ sec}}$$
the time taken to 
