Ian: New page: Using the equation $s=ut+\frac{1}{2}a{{t}^{\ 2}}$ gives $\begin{align} & 162=36t+\frac{1}{2}\times \left( -3 \right){{t}^{\ 2}} \\ & \\ & 324=72t-3{{t}^{\ 2}} \\ & \...$

2010-04-10T12:12:24Z

New page: Using the equation <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives <math>\begin{align} & 162=36t+\frac{1}{2}\times \left( -3 \right){{t}^{\ 2}} \\ & \\ & 324=72t-3{{t}^{\ 2}} \\ & \...

New page

Using the equation <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives

<math>\begin{align}
& 162=36t+\frac{1}{2}\times \left( -3 \right){{t}^{\ 2}} \\
& \\
& 324=72t-3{{t}^{\ 2}} \\
& \\
& {{t}^{\ 2}}-24t+108=0 \\
\end{align}</math>

This is a quadratic equation for
<math>t</math>
and can be solved by factorisation as follows.

We get

<math>\left( t-18 \right)\left( t-6 \right)=0</math>

giving

<math>t=6\ \text{s}</math>
or
<math>t=18\ \text{s}</math>

The first solution is the correct one. The second solution is the time taken for the car to decelerate down to zero velocity and then move backwards until it reaches the point
<math>s=162\ \text{m}</math>
once again.

This explanation can be verified as one finds the velocity is positive when
<math>t=6\ \text{s}</math>
and negative when
<math>t=18\ \text{s}</math>.

Solution 6.9 - Revision history

Ian: New page: Using the equation s=ut+\frac{1}{2}a{{t}^{\ 2}} gives \begin{align} & 162=36t+\frac{1}{2}\times \left( -3 \right){{t}^{\ 2}} \\ & \\ & 324=72t-3{{t}^{\ 2}} \\ & \...

Ian: New page: Using the equation $s=ut+\frac{1}{2}a{{t}^{\ 2}}$ gives $\begin{align} & 162=36t+\frac{1}{2}\times \left( -3 \right){{t}^{\ 2}} \\ & \\ & 324=72t-3{{t}^{\ 2}} \\ & \...$