2. A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance

Question

2.	A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance

flexastexas · Answer

What formula do I use?

aaronq · Answer

use two of the kinematic formulas to eliminate acceleration (i.e. substitute one into the other), like when solving two simultaneous equations.

flexastexas · Answer

How would I know which to use?

aaronq · Answer

it doesn't matter which ones you use, as long as you use the information given

$v=v_0+at\rightarrow  a=\dfrac{(v-v_0)}{t}$

then plug this into: $v^2=v^2_0+2a\Delta x$


$v^2=v^2_0+\dfrac{2(v-v_0)}{t}\Delta x$

solve for t

flexastexas · Answer

What do I plug into what? My initial velocity is 10m/s and final is 40m/s over a distance of 125m

flexastexas · Answer

@aaronq

aaronq · Answer

is this the first time you've seen these equations?

$v_0$ is the initial velocity
$v$ is the final velocity
$t$ is time
$\Delta x$ is the displacement

thephysicsman · Answer

acceleration is the change in velocity. it's the change in velocity over the change in time 

$$\frac{ v-v_0 }{ \Delta~t } = a $$