1. ## Kinematics-Constant Acceleration

I do not understand how the question can be solved if I have only been given the times and distances. I thought that you need 3 items.

There are two items to work out so how do I do this with only the distance and time:
P, Q and R are three points on a straight road, such that PQ=184m and QR=285.75m . A car traveling with constant deceleration passed points P, Q and R at times t=0, t=8 and t=25 seconds respectively. What is the retardation and the velocity of the car at P.

I know the equations for constant acceleration are:
v=u+at
v^2=u^2+2as
s=0.5(u+v)t
s=ut+0.5 * a * t^2

I tried working out the velocity using distance divided by time but this is average and not useful. I then tried rearranging the equation so a was the subject and then put the two equal to each other which did not work.

2. Originally Posted by aqua
I do not understand how the question can be solved if I have only been given the times and distances. I thought that you need 3 items.

There are two items to work out so how do I do this with only the distance and time:
P, Q and R are three points on a straight road, such that PQ=184m and QR=285.75m . A car traveling with constant deceleration passed points P, Q and R at times t=0, t=8 and t=25 seconds respectively. What is the retardation and the velocity of the car at P.

I know the equations for constant acceleration are:
v=u+at
v^2=u^2+2as
s=0.5(u+v)t
s=ut+0.5 * a * t^2

I tried working out the velocity using distance divided by time but this is average and not useful. I then tried rearranging the equation so a was the subject and then put the two equal to each other which did not work.
Using the kinematic equations gives

$\displaystyle x=x_0+v_0t+\frac{a}{2}t^2$

you can start at the origin this gives $x_0=0$

Now you know two other points $(x,t)$ that is
$(184,8)$ and $(469.75,25)$

This gives the system of equations
$\begin{array}{rcrcr} 8v_0 & +&32a & =&184 \\ 25v_0 & +& \frac{625}{2}a& =& 469.75\end{array}$

Can you finish from here