Question

A truck covers 40.0 m in 9.45 s while uniformly slowing down to a final velocity of 2.10 m/s.(a) Find the truck's original speed._____ m/s(b) Find its acceleration._____ m/s2

Answer 1

(a)

In order to find the original speed, let's use the formula below to find an expression for the acceleration:

`\begin{gathered} V=V_0+a\cdot t\\ \\ 2.1=V_0+a\cdot9.45\\ \\ a=(2.1-V_0)/(9.45) \end{gathered}`

Now, we can use the following formula to find the initial speed:

`\begin{gathered} \Delta S=V_0t+(at^2)/(2)\\ \\ 40=V_0\cdot9.45+(((2.1-V_0))/(9.45)\cdot9.45^2)/(2)\\ \\ 40=9.45V_0+4.725(2.1-V_0)\\ \\ 40=9.45V_0+9.9225-4.725V_0\\ \\ 4.725V_0=40-9.9225\\ \\ V_0=(30.0775)/(4.725)\\ \\ V_0=6.3656\text{ m/s} \end{gathered}`

(b)

Now, calculating the acceleration, we have: