You are travelling down a highway at 110km/h when you notice a turn up ahead that you will need to slow down for. You apply your breaks, which you know causes an acceleration of 2.15m/s^2 for 6.8s. Determine the final velocity
7 years ago
Answered By Jules E
First, you use the formula $a=\frac{v_f-v_i}{t}$a=vƒ−vit . Manipulate the formula to solve for final velocity:
$v_f=at+v_i$vƒ=at+vi . Since your given initial velocity is in km/h, you'll have to convert it to m/s this way:
7 years ago
Answered By Jules E
First, you use the formula $a=\frac{v_f-v_i}{t}$a=vƒ −vit . Manipulate the formula to solve for final velocity:
$v_f=at+v_i$vƒ =at+vi . Since your given initial velocity is in km/h, you'll have to convert it to m/s this way:
$\frac{110km}{1h}\times\frac{1h}{3600s}\times\frac{1000m}{1km}=$110km1h ×1h3600s ×1000m1km = 30.6m/s.
Then you simply place the given variables into the formula. Note that acceleration is negative because the car is slowing down.
$v_f=$vƒ = (-2.15m/s2)(6.8s)+30.6m/s
$v_f=$vƒ =16m/s