# Math Help - kinematics

1. ## kinematics

a car comes to a stop from a speed of 30m/s in a distance of 804m. the driver brakes so as to produce a deceleration of 1/2 m/s/s to begin with, then brakes harder to produce a deceleration of 3/2 m/s/s. Find the speed of the car at the instant the decelaration is increased, and the total time it takes for the car to stop.

i've toiled over this for a whole day. im sure there's something im missing! i dont understand how i can get v without t! so i drew up a graph and tried to combine all these equations but there are just too many unknowns for me! can someone help me solve this step by step?

2. Hello furor celtica
Originally Posted by furor celtica
a car comes to a stop from a speed of 30m/s in a distance of 804m. the driver brakes so as to produce a deceleration of 1/2 m/s/s to begin with, then brakes harder to produce a deceleration of 3/2 m/s/s. Find the speed of the car at the instant the decelaration is increased, and the total time it takes for the car to stop.

i've toiled over this for a whole day. im sure there's something im missing! i dont understand how i can get v without t! so i drew up a graph and tried to combine all these equations but there are just too many unknowns for me! can someone help me solve this step by step?
Suppose that the speed of the car when the deceleration changes is $v$, and that the distance travelled at this point is $s$. Then using:
$v^2=u^2+2as$
we get, for the first part of the motion:
$v^2 = 30^2-s$ (1)
and for the second part
$0=v^2-3(804-s)$
$=900-s -2412+3s$, from (1)
$\Rightarrow 2s = 1512$

$\Rightarrow s = 756$

$\Rightarrow v^2=900-756$

$\Rightarrow v = 12$
So the speed when the deceleration changes is $12$ m/s, and it takes $(30-12)\div\frac12=36$ seconds for the first part and $12\div\frac32=8$ seconds for the second part; total $44$ secs.