In this question we will work through the steps to solve this question:

A bus is travelling along a straight road at $25$ m$/$s $(90$ km$/$h$)$ when the driver sees a deer on the road and slams on the breaks. This causes a constant deceleration $($a negative acceleration$)$ of k$$ m$/$s$2.$ It takes the bus $5$ seconds to stop. How far does the bus travel before coming to a stop?

Steps in solution:

Let t$=0$ at the moment the bus driver slams on the breaks.

Let v$($t$)$ be the velocity of the bus at time t$($measured in m$/$s$).$

Let s$($t$)$ be the position of the bus at time t $($measured in meters$),$ and let's choose s$(0)=0.$

$$

a$)$ Find v$(0)$ in m$/$s$.[$ Select $]["$v$(0)=25","$v$(0)=90","$v$(0)=5","$v$(0)=0","$v$(0)=$ k$"]$

b$)$ What does "a constant deceleration $($a negative acceleration$)$ of k m$/$s$2"$ mean? $[$ Select $]["$v$\text{'}($t$)=-$ k$","$v$\text{'}($t$)=-5$k$","$s$($t$)=-5$k$","$s$($t$)=-$ k$","$s$\text{'}($t$)=-5$k$","$s$\text{'}($t$)=-$ k$"]$

c$)$ What does $"$It takes the bus $5$ seconds to stop" mean? $[$ Select $]["$v$(5)=$ s$(5)","$v$(5)=0","$s$(5)=0","$v$\text{'}(5)=0","$v$(0)=0"]$

d$)$ Use the information in parts $($a$)-($c$)$ to find a value for k$,$ and an expression for v$($t$)[$ Select $]["$k$=3$ and v$($t$)=-3$t$+90","$k$=25$ and v$($t$)=25$t $-5","$k$=5$ and v$($t$)=-25$t $-5","$k$=25$ and v$($t$)=-5$t$+25","$k$=5$ and v$($t$)=-5$t$+25"]$

e$)$ Use the information from $($d$)$ as well as s$(0)=0$to find an expression for s$($t$).[$ Select $]["$s$($t$)=25/2$ t$^2-5$ t$","$s$($t$)=-5/2$ t$^2-5$t$","$s$($t$)=-5/2$ t$^2+25$t$","$s$($t$)=-5$ t$^2+25$t$","$s$($t$)=-5$ t$^2+25$ t $+5"]$

f$)$ What should you evaluate to determine how far the bus travels before coming to a stop? $[$ Select $]["$s$(5)","$v$(25)","$v$(5)","$s$(0)","$v$(0)"]$

g$)$ Now find the final answer: How far does the bus travel before coming to a stop? $[$ Select $]["70\mathrm{.}5$ meters", $"25$ meters", $"0$ meters", $"62\mathrm{.}5$ meters", $"52\mathrm{.}5$ meters"$]$